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The movie Titanic broke box-office records by bringing in over $1,270,000,000 in worldwide distribution. (a) Express this amount in gigadollars. 1.27 gigadollars (1,270,000,000)e-9 (b) Express this amount in teradollars. .00127 teradollars (1,270,000,000)e-12
Velocity is related to acceleration and distance by the following expression, v^2 = 2ax^p. Find the power p that makes this equation dimensionally consistent. p = 1 Acceleration is the rate of change of velocity. If you know calculus, acceleration is the first derivative of velocity. If you don't know calculus, acceleration is the slope of the velocity curve or graph. All these definitions are equivalent.
The time T required for one complete oscillation of a mass m on a spring of force constant k is given by the formula below. T=2(pi)*sq(m/k) Find the dimensions k must have for this equation to be dimensionally correct. m/(t^2) Cancel 2pi, and solve for k
A vector A has a length of 5.7 m and points in the negative x direction. (a) Find the x component of the vector -5.3. 30.21 m (5.7)(5.3) (b) Find the magnitude of the vector -5.3 30.21 m (5.7)(5.3)
Use unit vector notation to express each of the vectors in Figure 3-39, in which the magnitudes of A, B, C, and D are respectively given by 33 m, 44 m, 22 m, and 33 m. A=(25.279466 x + 21.2119 y) m B=(41.602 x + -14.32499 y) m C=(-19.938 x + 9.297 y) m D=(0 x + 33 y) m Use formulas: X=mag(cosTHETA) Y=mag(sinTHETA)
The vector A in the drawing has a magnitude of 750 units. (a) Calculate the x and y components of the vector A, relative to the black axes. Ax = 649.5190 units 750(cos30) Ay = 375 units 750(sin30) (b) Calculate the x and y components of the vector A, relative to the colored axes. Ax' = 574.53 units 750(cos40) Ay' = -482.09 units -750(sin40)
The two tennis players walk to the net to congratulate one another, as shown shown in Figure 2-25, in which d1 = 5m and d2 = 2 m. (a) Find the distance traveled and the displacement of player A. 5 m (distance traveled) 5 m (displacement) (b) Repeat for player B. 2 m (distance traveled) -2 m (displacement) In the west (negative) direction, displacement is negative
In Figure 2-24, where d1 = 0.80 mi and d2 = 0.55 mi, you walk from the park to your friend's house, then back to your house. (a) What is your distance traveled? 2.6 mi .55+.35+.35+.55+.8 (b) What is your displacement? -.8 mi .8 in the negative direction
It was a dark and stormy night, when suddenly you saw a flash of lightning. Three-and-a-half seconds later you heard the thunder.Given that the speed of sound in air is about 340 m/s , how far away was the lightning bolt? 1190 m 340 x 3.5 = 1190m
You jog at 6.8 mi/h for 5.0 mi, then you jump into a car and drive for another 5.0 mi. With what average speed must you drive if your average speed for the entire 10.0 miles is to be 10.8 mi/h? 26.18 mi/hr Average speed = total distance ÷ total time 10.8 = 10.0 ÷ total time Total time = 0.926 hour Jog Speed = distance ÷ time 6.8 = 5 ÷ t Time = 0.735 hour Time of driving = 0.926 – 0.735 = 0.191 hr Distance driving = 5 mi Average speed = 5 ÷ 0.191 = 26.18 mi/hr
The position of a particle as a function of time is given by x = (2.4 m/s)t + (-2.8 m/s2)t2. (a) Plot x-versus-t for time from t = 0 to t = 1.0 s. (Do this on paper. Your instructor may ask you to turn in this plot.) (b) Find the average velocity of the particle from t = 0.45 s to t = 0.55 s. -.4 m/s v(t)=dx(t)/dt v(t) = 2.4 -2*2.8*t v(0.45)=-0.12 v(0.55)=-0.68 Average = -0.4 (c) Find the average velocity from t = 0.49 s to t = 0.51 s. -.4 m/s v(0.49)=-0..344 v(0.51)=-0.456 Average = -0.4
A jet makes a landing traveling due east with a speed of 110 m/s. If the jet comes to rest in 12.5 s, what is the magnitude and direction of its average acceleration? -8.8 m/s due west v=at+vo a=(v-vo)/t a=(0-110)/12.5 = 8.8
A motorcycle moves according to the velocity-versus-time graph shown in Figure 2-31. (The vertical axis is marked in increments of 6 m/s and the horizontal axis is marked in increments of 7 s.) Find the average acceleration of the motorcycle during each of the segments of motion, A, B, and C. aA = 1.7142857 m/s2 [Velocity at A]-{Velocity at B] / [Time at A]-{Time at B] = (12-0)/(7-0) aB = 0 m/s2 (12-12)/(21-14) aC = -.4285714 m/s2 (6-12)/(35-21)
When you see a traffic light turn red, you apply the brakes until you come to a stop. If your initial speed was 13 m/s, and you were heading due west, what was your average velocity during braking? Assume constant deceleration. Magnitude: 5.5 m/s (11-0)/(2-0) Direction: West
When you see a traffic light turn red, you apply the brakes until you come to a stop. Suppose your initial speed was 12.9 m/s, and you come to rest in 34.5 m. How much time does this take? Assume constant deceleration. vf=0, v0=12.9 and d=34.5, so 0^2=12.9^2 +2*a*34.5 => a= - 2.4m/s/s then, t=(0-12.9)/-2.4 = 5.3s
A child slides down a hill on a toboggan with an acceleration of 1.6 m/s^2. If she starts at rest, how far has she traveled after 2, 4, and 6 seconds? d= v0*t + 1/2 a*t^2 since v0=0. then d= 1/2 a*t^2 t=2 then d=3.2 m t=4 then d=12.8 m t=6 then d=28.8 m
On a ride at an amusement park, passengers accelerate straight downward from zero to 46.5 mi/h in 3.2 seconds. What is the magnitude of the average acceleration of the passengers on this ride? First, convert 46.5mi/hr to m/s: (46.5*1609.34m)/3600s = 20.787 v=at+vo 20.787=3.2a+0 6.478515625 m/s2
Referring to the cartoon below, how long would it take for the car to go from 0 to 2 mi/h? convert m/hr to ft/s 2 m/h=2 miles*(5280ft/mile)/((1hr*360sec/hr) = 29.33 ft/sec v=at+vo (vo is 0) at=v so t=v/a = (29.33 ft/sec)/(32.2 ft/sec2) = 0.91 sec
Seagulls are often observed dropping clams and other shellfish from a height to the rocks below, as a means of opening the shells. If a seagull drops a shell from rest at a height of 19 m, how fast is the shell moving when it hits the rocks? V^2= Vo^2 + 2ax V= sqrt(0+ 2 * 9.81m/s^2 * 19m) V=19.3 m/s
You are driving up a long inclined road. After 1.7 miles you notice that signs along the roadside indicate that your elevation has increased by 520 ft. (a) What is the angle of the road above the horizontal? Convert 1.7miles to feet 1mile = 5280 feet 1.7 miles x 5280 feet/1mile = 8976 feet = Distance Elevation = 520 feet sin <a = A / B sin <a =520 / 8976 = 3.321 (b) How far do you have to drive to gain an additional 150 ft of elevation? Additional 150 ft elevation: 150 ft + 520 ft = 670 ft Distance = A = 8976 + x (x symbolizes the additional distance traveled) sin 3.321 = 670 / 8976 + x 0.0579(8976 + x) = 670 519.979 + 0.0579x = 670 0.0579x = 670 - 519.979 0.0579x/0.0579 = 150.02/0.0579 x = 2591.03 ft
(a) Find the x and y components of a position vector, r, of magnitude r = 75 m, if its angle relative to the +x axis is 25.0°. A.) (x component) B.) (y component) x - component = 75 cos(25) = 75 x 0.9063 = 38 m (nearly) y - component = 75 sin(25) = 75 x0.4226 = 32 m (nearly)
(b) What if the relative angle is instead 65.0°? C.) (x component) D.) (y component) x - component = 75 cos(65) = 75 x0.4226 = 32 m (nearly) y - component = 75 sin(65) = 75 x0.0.9063 = 68 m ( nearly)
A vector "A" has a magnitude of 40.0 and points in a direction 20.0 below the positive axis. A second vector, has a magnitude of 75.0 and points in a direction 50.0 above the positive axis.a) Sketch the vectors A, B, and C = A + B. (b) Using the component method of vector addition, find the magnitude and direction of the vector C. Ax= 40cos20° = 37.6 Ay= - 40sin20° = -13.7 Bx= 75cos50°= 48.2 By= 75sin50° = 57.5 Cx= Ax + Bx = 37.6 + 48.2 = 85.8 Cy = Ay + BY = 57.5 - 13.7 = 43.8 magnitude of C = √(Cx)² + (Cy)² = 96.33 Direction of C = θ = arctan(Cy/Cx) = 27.04
Vector A points in the negative x direction and has a magnitude of 22 units. The vector B points in the positive y direction. (a) Find the magnitude of B if A + B has a magnitude of 39 units. 32.202 (b) Sketch A and B. (Do this on paper. Your instructor may ask you to turn in this diagram.)
A vector has a magnitude of 2.4 m and points in a direction that is 150° counterclockwise from the x axis. Find the x and y components of this vector. Ax= -2.0784 m 2.4(cos150) Ay= 1.2 m 2.4(sin150)
The vector -5.7A has a magnitude of 40 m and points in the positive x direction. (a) Find the x component of the vector . -7.017 m 40/-5.7 (b) Find the magnitude of the vector 7.017 m 40/5.7
If A = (29 m) x + (-12 m) y, and B = (2.0 m) x + (14 m) y, find the direction and magnitude of each of the following vectors.(a) A 337.5205 ° (counterclockwise from the +x-axis) arctan(-12/29) + 360 31.3847 m sq(29^2 + 12^2)
(b) B 81.8698 ° (counterclockwise from the +x-axis) arctan(14/2) 14.1421 m sq(2^2 + 14^2)
(c) A + B 3.6913 ° (counterclockwise from the +x-axis) arctan[ (-12+14)/(29+2) ] 31.06444 m sq[ (29+2)^2 + (-12+14)^2 ]
Two of the allowed chess moves for a knight are shown in Figure 3-40. (a) Is the magnitude of displacement 1 greater than, less than, or equal to the magnitude of displacement 2? equal to (b) Find the magnitude and direction of the knight's displacement for each of the two moves. Assume that the checkerboard squares are 3.46 cm on a side. Magnitude 1 7.7367 cm sq[ (3.46)^2 + (2*3.46)^2 ] Direction 1 153.4349° (counterclockwise from the x-axis) arctan(6.92/3.46) + 90 Magnitude 2 7.7367 cm sq[ (3.46)^2 + (2*3.46)^2 ] Direction 2 63.4349° (counterclockwise from the x-axis) arctan(6.92/3.46)
In its daily prowl of the neighborhood, a cat makes a displacement of 127 m due north, followed by a 65 m displacement due west. Find the magnitude and direction of the displacement required for the cat to return home. 142.667 m sq( 127^2 + 65^2 ) 62.8961 ° south of east arctan(127/65)
As an airplane taxies on the runway with a speed of 16.5 m/s, a flight attendant walks toward the tail of the plane with a speed of 1.22 m/s. What is the flight attendant's speed relative to the ground? 15.3 m/s (16.5-1.22)
You are riding in a boat whose speed relative to the water is 6.1 m/s. The boat points at an angle of 25 degrees upstream on a river flowing at 1.4 m/s. Find the time it takes for the boat to reach the opposite shore if the river is 35 m wide. Known: vbw = 6.1 and d = 35 use formula v= x/t solve for t t= x/v where v is 6.1 cos 25 t= 35/6.1cos25 t= 6.33 s
A sailboat runs before the wind with a constant speed of 4.4 m/s in a direction 37 degrees north of west. (a) How far west has the sailboat traveled in 21 min? (km) speed West= 4.4cos37 Distance West=speed*time=4.4cos37(1680s)= 5903.5 m (b) How far north has the sailboat traveled in 21 min? (km) speed North= 4.4sin37 distance North=speed*time=4.4sin37(1680s)= 4448.6 m
As you walk to class with a constant speed of 1.20 m/s you are moving in a direction that is 15.0° north of east. (a) How much time does it take to change your displacement by 26.0 m east? east component of speed = 1.2 cos (15) = 1.16m/s time east = 26/1.16 = 22.4 s (b) How much time does it take to change your displacement by 24.0 m north? north component of speed = 1.2 sin (15) = 0.31m/s time north = 24/.31 = 77.4 s
An archer shoots an arrow horizontally at a target 15 m away. The arrow is aimed directly at the center of the target, but it hits 48 cm lower. What was the initial speed of the arrow? (Neglect air resistance.) Find flight time: 0.48 = (1/2) * g * t^2 t= sqr(2*0.48/9.8) = 0.31298 s
vx = x/t = 15/0.31298 = 47.926 m/s vy = g*t = -9.8*0.31298 = -3.067 m/s
Initial speed of the arrow: v = sqr(vx^2 + vy^2) v = sqr(47.926^2+( -3.067^2)) = 48.02 m /s
An astronaut on the planet Zircon tosses a rock horizontally with a speed of 6.95m/s. The rock falls through a vertical distance of 1.40m and lands horizontal distance of 8.75m from the astronaut. What is the acceleration of gravity on Zircon The horizontal equation is: x = Vt (where x is distance, V is horizontal velocity, t is time of flight) 8.75 = 6.95t t = 1.259 seconds
The vertical displacement equation is: y = (1/2)at^2 (where a is vertical acceleration) 1.4 = (1/2)a(1.259 )^2 1.4 (2)/(1.259 )^2= a a = 1.77 m/s/s
A soccer ball is kicked with a speed of 9.85 m/s at an angle of 45 degrees above the horizontal.? If the ball lands at the same level from which it was kicked, how long was it in the air? Distance traveled by the ball is given by the equation d = v^2 sin(2theta)/g 9.85^2 *sin90/9.81 = 9.89m, (where g - acceleration due to gravity = 9.81 m/s^2) Time of flight is given by t = d/v cos(theta) = 9.89/[9.85 * cos45] t = 9.89/ 6.965 = 1.42 s
A dolphin jumps with an initial velocity of 11.5 m/s at an angle of 32.0° above the horizontal. The dolphin passes through the center of a hoop before returning to the water. If the dolphin is moving horizontally when it goes through the hoop, how high above the water is the center of the hoop? The y component of the dolphin's initial velocity is equal to (11.5)sin32 = 6.095 m/s. Use the equation vfinal^2 = vinitial^2 + 2a(change in y). (0)^2 = (6.095)^2 + (2)(-9.8)(change in y) 0 = 37.15 - 19.6(change in y) change in y = 1.895 m
A golf ball is struck with a five iron on level ground. It lands 105 away 3.80 later. What was (a) the direction and (b) the magnitude of the initial velocity? The horizontal velocity is constant during the flight since there are no horizontal forces acting; Vhorizontal = 105m/3.8s = 27.6m/s If the ball is in flight for 3.8s, it took 1.9s to reach max alt, this means that its initial vertical velocity is 9.8m/s/s x 1.9s = 18.6m/s so initial launch angle is given by tan(theta) = 18.6/27.6 => theta = 34 deg initial speed sqrt[18.6^2+27.6^2] = 33.3m/s
An astronaut on Uranus drops a rock straight downward from a height of 0.75 m. If the acceleration of gravity on Uranus is 8.93 m/s2, what is the speed of the rock just before it lands? V = √[2*a*d] = √[2*8.93*.75] = 3.66 m/s
You slide a box up a loading ramp that is 13.0 ft long. At the top of the ramp the box has risen a height of 3.20 ft. What is the angle of the ramp above the horizontal? 14.2500 °
A baseball player slides into third base with an initial speed of 8.70 m/s. If the coefficient of kinetic friction between the player and the ground is 0.37, how far does the player slide before coming to rest? The force slowing the player down is F = -0.37*m*g = -3.63*m So the player's acceleration is a = F / m = -3.63 m/s^2 (m cancels) Now use v^2 = vo^2 + 2ad v = 0, vo = 8.7m/s, a = -3.63m/s^2 so distance of slide is d = -8.7^2 / (2*-3.63) m = 10.4 m
Hopping into your Porsche, you floor it and accelerate at 12.9 m/s2 without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible. Newton's Third Law: To every action there is an equal and opposite reaction. Which means the force of acceleration F = ma should be equal to the force of friction between the road and tire. Force of friction is F = f x Fn where f = coeff of friction, Fn = normal force = mg Putting F = m a and F = f m g together.. you get: m a = f m g Therefore, the minimum coeff of friction you require is f = a/g = 12.9/9.81 1.315
To move a large crate across a rough floor?, you push on it with a force F at an angle of 21°, below the horizontal, as shown in the figure. Find the acceleration of the crate, given that the mass of the crate is m = 29 kg, the applied force is 339 N and the coefficient of kinetic friction between the crate and the floor is 0.51 By F(net) = F(applied) - F(friction) =>F(net) = F x cosθ - µk x N =>F(net) = Fcosθ - µk x [mg + Fsinθ] =>m x a = 339 x cos21* - 0.51 x [29 x 9.8 + 339 x sin21*] => a = 109.58/29 =>a = 3.78 m/s^2
To move a large crate across a rough floor, you push on it with a force F at an angle of 27° below the horizontal, as shown in Figure 6-21. Find the force necessary to start the crate moving, given that the mass of the crate is m = 34 kg and the coefficient of static friction between the crate and the floor is 0.54. The horizontal fraction of the force is = F*cos27 The vertical fraction of the force is = F*sin27 The friction force by the crate alone is μmg = 0.54*34*9.81 N The friction force by the vertical component of the force on the crate is= μF*sin27
To make the crate move, the horizontal component of the force must at least be equal to the sum of the friction forces: 0.54*34*9.81 + 0.54*F*sin27 = F*cos27 F(cos27 - 0.54*sin27) = 0.54*34*9.81 F = 0.54*34*9.81/(cos27 - 0.54*sin27) F = 278.88 N
Pulling down on a rope, you lift a 4.40 kg bucket of water from a well with an acceleration of 2.50 m/s2. What is the tension in the rope? sumF = ma T-mg = ma T=m(a+g) T=4.40(2.50+9.81) T=54.164 N
A 48 kg person takes a nap in a (lightweight) backyard hammock. Both ropes supporting the hammock are at an angle of 21° above the horizontal. Find the tension in the ropes. sumF(y) = ma sumF(y) = 0 (there is no acceleration) Tsin(21)+Tsin(21)-mg = 0 2(Tsin21) = mg T = (.5mg)/sin21) T = 656.978 N
Your friend's 4.0 g graduation tassel hangs on a string from his rear-view mirror.(a) When he accelerates from a stoplight, the tassel deflects backward toward the rear of the car hanging at an angle of 6.3° relative to the vertical. Find the tension in the string holding the tassel. 1.4 g = 0.0014 kg mg = 0.0014 x 9.8 = 0.01372 N 90 - 6.3 = 83.7 degrees from horizontal 0.01372/sin(83.7) =0.0138 N (b) At what angle to the vertical will the tension in the string be twice the weight of the tassel? mg = 0.01372 N arcsin(0.01372/(2 x 0.01372)) = 30 degrees from horizontal 90 - 30 = 60 degrees from vertical
In Figure 6-26 we see two blocks connected by a string and tied to a wall, with θ = 31°. The mass of the lower block is m = .8 kg; the mass of the upper block is 2.0 kg. (a) Find the tension in the string connecting the two blocks. 4.038N m*g*sin(31) .8*9.8*sin(31) (b) Find the tension in the string that is tied to the wall. 14.147N (m1+m2)*g*sin31 (2.8)*9.8*sin31
Two blocks are connected by a string, as shown in Figure 6-28. The smooth inclined surface makes an angle of θ = 42° with the horizontal, and the block on the incline has a mass m1 = 6.7 kg. Find the mass m2 of the hanging block that will cause the system to be in equilibrium. m2*g = m1*g*sin42 m2 = 6.7*sin(42) 4.483N
You want to nail a 1.6 kg board onto the wall of a barn. To position the board before nailing, you push it against the wall with a horizontal force F to keep it from sliding to the ground. (a) If the coefficient of static friction between the board and the wall is 0.79, what is the least force you can apply and still hold the board in place? sumF = 0 mewN = mg N = (mg)/mew N = (15.696)/.79 = 19.868N (b) What happens to the force of static friction if you push against the wall with a force greater than that found in part (a)? Stays the same. An increase in normal force, doesn't affect the coefficient of friction.
A force of 9.4 N pulls horizontally on a 1.1 kg block that slides on a rough, horizontal surface. This block is connected by a horizontal string to a second block of mass m2 = 2.10 kg on the same surface. The coefficient of kinetic friction is µk = 0.22 for both blocks.(a) What is the acceleration of the blocks? Force of friction = µgm1 + µgm2 = 2.3716 + 4.5276 = 6.9 N a = Fnet/Σm = (Fa - Ff)/Σm = (9.4 - 6.9)/(1.1+2.1) = .78125 m/s² (b) What is the tension in the string? T = m2(a + µg) = 2.1(.78125 + .22*9.8) = 6.17 N
A child goes down a playground slide that is inclined at an angle of 25.0° below the horizontal. Find the acceleration of the child given that the coefficient of kinetic friction between the child and the slide is 0.424. ΣF(x) = ma = mgsinθ – μmgcosθ a = g(sinθ – μcosθ) = 9.81m/s²(sin24.0° - 0.425cos24.0°) = 0.181m/s²
Biomechanical research has shown that when a 67 kg person is running, the force exerted on each foot as it strikes the ground can be as great as 2300 N. (a) What is the ratio of the force exerted on the foot by the ground to the person's body weight? 2300/(67*9.81) = 3.499 (b) If the only forces acting on the person are (i) the force exerted by the ground and (ii) the person's weight, what is the magnitude and direction of the person's acceleration? Net Force: (2300)-(67*9.81) = 1642.73N F=ma; 1642.73=ma 24.52 m/s² = a Upward (c) If the acceleration found in part (b) acts for 10.2 ms, what is the resulting change in the vertical component of the The net force Fn = m*a = 67*24.52 = 1642.84 N Impulse P = Fn*t = 1642.84*.0102 = 16.75 N∙s dV = P/m = 16.75/67 = .25 m/s
An object of mass m = 5.95 kg has an acceleration a = (1.17 m/s/s)x + (-0.664 m/s/s)y. Three forces act on this object: F1, F2, and F3. Given that F1 = (3.22 N) x and F2 = (-1.55 N) x + (2.05 N) y, what is F3? the net force is F = m a = 5.95 [ 1.17 i - 0.664 j ] = 6.96 i - 3.95 j but net force F = F1 + F2 + F3 ie., F3 = F - F1 - F2 = 6.96 i - 3.95 j - 3.22 i + 1..5 i - 2.05 j = 5.29 i - 6.4 j
Find the tension in each of the strings in Figure 6-32, given that m1 = 1.4 kg, m2 = 1.9 kg, and m3 = 3.4 kg. Assume the table is frictionless and the masses move freely. acceleration = F/mtotal = gm3/(m1+m2+m3) = 4.973 m/s/s string between m1 and m2 F=ma; F=(4.973)*m1 = 6.962N string between m2 and m3 F=ma; F=(4.973)*(m1+m2) = 16.41N
Two blocks are connected by a string, as shown in Figure 6-33. The smooth inclined surface makes an angle of θ = 35° with the horizontal, and the block on the incline has a mass of 5.7 kg. The mass of the hanging block is 2.5 kg. Find the magnitude and direction of the hanging block's acceleration. ΣF = Ma let m1 = mass on the incline (5.7 kg), m2 = hanging mass (2.5 kg) m1*g*sinθ - m2*g = (m1 + m2)a (5.7)(9.8)(sin35) - 2.5(9.8) = (5.7+2.5)a solve for a 0.9195 m/s/s
Find the acceleration of the masses shown in Figure 6-32, given that m1 = 1.2 kg, m2 = 2.0 kg, and m3 = 3.5 kg. Assume the table is frictionless and the masses move freely. a = F/m a = (m3*9.81)/(3.5+2.0+1.2) a = 5.125
When you take your 1200 kg car out for a spin, you go around a corner of radius 55.8 m with a speed of 15.3 m/s. The coefficient of static friction between the car and the road is 0.94. Assuming your car doesn't skid, what is the force exerted on it by static friction? Centripetal force is mv²/R= (1200)(15.3)^2 / 55.8 =
5034N (5.034 kN) Friction is mgµ=11054N Force exerted by friction is the centripetal force 5034N. Friction calculated as mgµ represents the maximum possible value.
A car goes around a curve on a road that is banked at an angle of 32.0°. Even though the road is slick, the car will stay on the road without any friction between its tires and the road when its speed is 25.0 m/s. What is the radius of the curve? Vertical component of normal force=mg (weight of car) Horizontal component is centripetal force mv²/R Normal force is 32° from vertical. tan32°=(mv²/mgR)=(v²/gR) R=v²/(g*tan32°) = (25)^2 / (9.8*tan32) = 102.062m
Driving in your car with a constant speed of 12 m/s, you encounter a bump in the road that has a circular cross section, as indicated in Figure 6-35. If the radius of curvature of the bump is r = 37 m, find the apparent weight of a 74 kg person in your car as you pass over the top of the bump. m g - N = m v² / R N = mg - m v² / R But N is just the apparent weight of the person, so N = m (g - v²/R) = 74 (9.8 - 12^2/37) = 437.2N
You swing a 4.5 kg bucket of water in a vertical circle of radius 5.0 m.(a) What speed must the bucket have if it is to complete the circle without spilling any water? Fnet = ma mg + N = mv²/r Minimum speed occurs when N = 0: g = v²/r v = srqt(gr) = srqt(9.8*5) v = 7 (b) How does your answer depend on the mass of the bucket? the speed is independent of the mass
A 0.075 kg toy airplane is tied to the ceiling with a string. When the airplane's motor is started, it moves with a constant speed of 1.19 m/s in a horizontal circle of radius 0.43 m, as illustrated in Figure 6-40. Find the angle the string makes with the vertical and the tension in the string. the forces balance in the vertical direction T cosθ = mg (1) T sinθ = mv^2 / r (2) Dividing (1) and (2) : tan θ = v^2 / rg = 1.19^2/(9.8*.43) θ = 18.57°
From (1) T = mg / cos θ = 0.775 N
A hockey puck of mass m is attached to a string that passes through a hole in the center of a table, as shown in Figure 6-46. The hockey puck moves in a circle of radius r. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest? (Use g for acceleration due to gravity, and m, r, and M as necessary.) ((g*M*r)/m)^(1/2) (Same answer for everybody)
The gardening tool shown in Figure 11-24 is used to pull weeds. If a 1.24 N·m torque (about the pivot point) is required to pull a given weed, what force did the weed exert on the tool? (Figure shows distance of .04m) 31N F=(1.24)/.04
A person slowly lowers a 3.9 kg crab trap over the side of a dock, as shown in Figure 11-25. What torque does the trap exert about the person's shoulder? (Figure shows radius of .7) torque = F*d = 3.9kg*9.8m/s^2*0.70m = 26.8 N-m
A candy bar has a mass of 0.048 kg and a calorie rating of 205 Cal. What speed would this candy bar have if its kinetic energy were equal to its metabolic energy. [Note: The nutritional calorie, 1 Cal, is equivalent to 1000 calories (1000 cal) as defined in physics. In addition, the conversion factor between calories and joules is as follows: 1 Cal = 1000 cal = 1 kcal = 4186 J.] Kinetic energy of the candy bar = 205 * 4186 J = 1/2 m v2 Therefore, it’s velocity v = sqrt ( 2*205*4186/0.048) = 5.98e3 m/s = 5.98 km/s
You push a 40 kg box across a floor where the coefficient of kinetic friction is µk = 0.55. The force you exert is horizontal. (a) How much power is needed to push the box at a speed of 0.70 m/s? 151 W (b) How much work do you do if you push the box for 44 s? (a) Power = Force dot_product velocity Force = Friction force = N * µk = mg *µk= 40 * 9.81 * 0.55 N Thus, power P = Force * speed * cos (angle between force and velocity) = 40*9.81*0.55*0.70 = 151 W (b) W = P * Δt = 151 W * 44s = 6.64 kJ
After a tornado, a 0.60 g straw was found embedded 2.7 cm into the trunk of a tree. If the average force exerted on the straw by the tree was 80 N, what was the speed of the straw when it hit the tree? 84.9 m/s The force exerted on the straw to slow the straw to rest. The average acceleration of the straw is a = F/m = 80/0.60e-3 = 1.33e5m/s2 The distance traveled the straw; S = vi2/2a = 2.7e-2m vi = sqrt (2 * a * S) = sqrt ( 2 * 1.33e5 * 2.7e-2) = 84.9 m/s
The water skier in Figure 7-16 is at an angle of 34° with respect to the center line of the boat, and is being pulled at a constant speed of 14 m/s. (a) If the tension in the tow rope is 94.0 N, how much work does the rope do on the skier in 10.0 s? 10.9 kJ (b) How much work does the resistive force of water do on the skier in the same time? -10.9 kJ (a) Work W = F * ΔX * cos(angle) = 94.0 * (14.0*10.0) * cos (34°) = 10.9 kJ (b) W = Δ K = 0 = Wrope + Wresist è Wresist = -10.9 kJ
The force shown in Figure 7-17 acts on an object that moves along the x axis. (a) How much work is done by the force as the object moves from x = 0 to x = 2.2 m? 1.08 J (b) How much work is done by the force as the object moves from x = 0.6 m to x = 4.0 m? 1.5 J (c) How much work is done by the force as the object moves from x = 2.0 m to x = 3.7 m?0.626 J Work = area underneath (a) 1.08 J (b) 1.5 J (c) 0.626 J
The motor of a ski boat produces a power of 36,600 W to maintain a constant speed of 15.0 m/s. To pull a water skier at the same constant speed, the motor must produce a power of 38000 W. What is the tension in the rope pulling the skier? 93.3 N <variables referred here are all scalar: the magnitudes) The extra power is to keep the skier in constant speed. W = (Presist + Pextra) *Δ t =Δ K = 0 è Presist= Fr * V * cos (angle) = - Pextra = -(38000-36600) = -1400W Fr = 93.3 N The skier is in constant velocity motion è T - Fr = 0 è T = 93.3 N
To make a batch of cookies you mix half a bag of chocolate chips into a bowl of cookie dough, exerting a 21.5 N force on the stirring spoon. Assume that your force is always in the direction of motion of the spoon. (a) What power is needed to move the spoon at a speed of 0.27 m/s? 5.81 W (b) How much work do you do if you stir the mixture for 2.0 min? 0.697 kJ (a) P = F * V * cos(angle) = 21.5 * 0.27 * cos(0) = 5.81 W (b) W = P *Δ t = 5.81 * 2*60 = 697 J = 0.697 kJ
A catapult launcher on an aircraft carrier accelerates a jet from rest to 71.9 m/s. The work done by the catapult during the launch is 6.5 107 J. (a) What is the mass of the jet? 25100 kg (b) If the jet is in contact with the catapult for 2.7 s, what is the power output of the catapult? 24.1 MW (a) W =Δ K = Kf = 1/2 m v2 è m = 2*W/v2 = 25100kg (b) P = W/Δt = 6.5e7 / 2.7 = 2.41e7 W = 24.1 Mw
1440 kg car delivers a constant 46.3 hp to the drive wheels. We assume the car is traveling on a level road and that all frictional forces may be ignored. (a) What is the acceleration of this car when its speed is 14.9 m/s?1.61 m/s2 (b) If the speed of the car is cut in half does its acceleration increase, decrease, or stay the same? increases (c) Calculate the car's acceleration when its speed is 26.6 m/s. 0.902 m/s2 (a) P = F * V *cos (angle) , when angle = 0, è P = F * V è F = P/V = m a è a = P/V/a = 46.3*746/14.9/1440 = 1.61m/s2 (b) increases, since a is inverse proportional to V (c) a = 46.3*746/26.6/1440 = 0.902 m/s2
A string that passes over a pulley has a 0.321 kg mass attached to one end and a 0.655 kg mass attached to the other end. The pulley, which is a disk of radius 9.50 cm, has friction in its axle. What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium? The force of the 0.321kg block is 0.321 * 9.81 (gravity) = 3.14901N The force of the 0.655kg block is 0.655 * 9.81 = 6.42555N Distance from centre of the pulley is 0.0475m (3.14901 * 0.0475) = .149577975 (6.42555 * 0.0475) = .30521125 0.30521125-0.149577975 =
0.155Nm
To determine the location of his center of mass, a physics student lies on a lightweight plank supported by two scales L = 2.50 m apart, as indicated in the figure below. (a) If the left scale reads 290 N, and the right scale reads 122 N, find the student's mass. The student's mass is: (290 + 122) / 9.81 = 42.0 kg (b) Find the distance from the student's head to his center of mass. 290x = 122(2.50 - x) 2.377x = 2.50 - x 3.377x = 2.50 x = .740 m
A school yard teeter totter with a total length of 5.1 m and a mass of 36 kg is pivoted at its center. A 25 kg child sits on one end of the teeter totter. (a) Where should a parent push with a force of 187 N in order to hold the teeter totter level? 25*9.8*(5.1/2)=187x x=3.34 m (from the pivot) (b) Where should the parent push with a force of 338 N to keep the teeter totter level? 25*9.8*(5.1/2)=338x x=1.84 m (from the pivot) (c) How would your answers to parts (a) and (b) change if the mass of the teeter totter were doubled? Explain. The change in mass of the teeter totter would not matter because the mass balances itself out.
A 0.13 kg meter stick is held perpendicular to a wall by a 2.7 m string going from the wall to the far end of the stick. (a) Find the tension in the string. Sin(a) = 1 / 2.7 a = 21.7385 degrees
The vertical component of the tension of the string = (0.13 / 2) * 9.81 = 0.63765 N
The tension in the string = 0.63765 / Cos(21.7385) = 0.6865 N (b) If a shorter string is used, will its tension be greater than, less than, or the same as that found in part (a)? greater than (c) Find the tension in a 1.6 m string. Sin(a) = 1 / 1.5 a = 41.81031 degrees
The vertical component of the tension of the string = (0.13 / 2) * 9.81 = 0.63765 N
The tension in the string = 0.63765 / Cos(41.81031) = 0.8555 N
A uniform metal rod, with a mass of 3.5 kg and a length of 1.3 m, is attached to a wall by a hinge at its base. A horizontal wire bolted to the wall 0.55 m above the base of the rod holds the rod at an angle of 25° above the horizontal. (a) Find the tension in the wire. mg = 3.5*9.8 = 34.3 T = [34.3 * 1.3/2 * cos(25)] / 0.55 36.74 N (b) Find the horizontal and vertical components of the force exerted on the rod by the hinge. Net horizontal force = 0. Therefore horizontal component of foce by the hinge = T = 36.74N Net vertical force = 0 Therefore vertical component of force by the hinge = mg = 34.3 N
A rigid, vertical rod of negligible mass is connected to the floor by an axle through its lower end, as shown in the figure below. The rod also has a wire connected between its top end and the floor. A horizontal force F is applied at the midpoint of the rod. (a) Find the tension in the wire in terms of F. (Answer without using any trig functions.) F/sqrt 2 (b) Find the horizontal and vertical components of force exerted by the bolt on the rod in terms of F. F(0.5) (horizontal component) F(0.5) (vertical component)
A uniform crate with a mass of 14.7 kg rests on a floor with a coefficient of static friction equal to 0.558. The crate is a uniform cube with sides 1.21 m in length.(a) What horizontal force applied to the top of the crate will initiate tipping? The weight of the cube (14.7kg*9.81m/s2 = 144.2 N) will be at a radius of (1.21m/2 = ) 0.61m. F*1.21 = 144.2*0.61 F = 72.7 N (b) If the horizontal force is applied halfway to the top of the crate it will begin to slip before it tips. Explain. Because the restorative force needed would be too high
In each hand you hold a 0.10 kg apple. (a) What is the gravitational force exerted by each apple on the other when their separation is 0.50 m? F = [(6.67e-11)*(.10)(.10) / (.5^2) ] F = 2.668e-12 N (b) What is the gravitational force exerted by each apple on the other when their separation is 1.30 m? F = [(6.67e-11)*(.10)(.10) / (1.3^2) ] F = 3.9467e-13
A 7.1 kg bowling ball and a 7.7 kg bowling ball rest on a rack 1.35 m apart. (a) What is the force of gravity exerted on each of the balls by the other ball? F = (6.67 x 10^-11) x (7.1 x 7.7) /(1.35)^2 = 2 x 10^-9 N (b) At what separation is the force of gravity between the balls equal to 2.0 * 10^-9 N? 2.0 e 10-9 = (6.67 x 10^-11) x (7.1 x 7.7) /r^2 r = 1.35m
The largest asteroid known, Ceres, has a mass of roughly 8.7 * 10^20 kg. If Ceres passes within 11,000 km of the spaceship in which you are traveling, what force does it exert on you? Use 72 kg as an approximate value for your mass. 6.673e-11*8.7e20*72/(11000e3)^2 = 0.035 N
Three 6.30 kg masses are at the corners of an equilateral triangle and located in space far from any other masses. F = 6.67 * 10^(-11) * 6.30^2 sqrt(3) / 1.90^2 = 1.27 * 10^(-9) N. (b) What would your answer to part (a) be if instead the sides of the triangle are doubled in length? It is evident from (1) that doubling D divides the force by 4. 1.27 * 10^(-9) N / 4
(a) Find the acceleration due to gravity on the surface of Mars. mMars=6.4191 × 10^23 kg rMars=3.376 × 10^6 m g = GM/r² g = 6.67300 × 10-11 * 6.4191 × 10^23 / (3.376 × 10^6)² g = 3.758 m/s² (b) Find the acceleration due to gravity on the surface of Uranus. *Different celestial bodies for everyone* Look up mass and radius of celestial body. Use equation g = GM/r²
Two 6.3 kg balls, each with a radius of 0.11 m, are in contact with one another. What is the gravitational attraction between the balls? F = GmM/r^2 r = 2(.11) = .22 F = (6.67x10^-11)(6.3)^2 / (0.22)^2 F = 5.4x10^-8 N
The acceleration due to gravity on the surface of the Moon is known to be about 1/6 the acceleration due to gravity on the Earth. Given that the radius of the Moon is roughly 1/4 that of the Earth, find the mass of the Moon in terms of the mass of the Earth. 0.0104 MEarth Should be the same answer for everybody. If not post the problem and solution to the BATS FORUM
Consider a spherical asteroid with a radius of 21 km and a mass of 8.7 *10^15 kg. Assume the asteroid is roughly spherical.(a) What is the acceleration due to gravity on the surface of this asteroid? 21km = 21,000m g=GM/r^2 g=(6.67e-11)(8.7e15)/(21000^2) g =1.316 * 10^-3 m/s² (b) Suppose the asteroid spins about an axis through its center, like the Earth, with a rotational period T. What is the smallest value T can have before loose rocks on the asteroid's equator begin to fly off the surface? ANSWERS NOT AVAILABLE, PLEASE VISIT FORUM
On Apollo missions to the Moon, the command module orbited at an altitude of 140 km above the lunar surface. How long did it take for the command module to complete one orbit? **look up mass and radius of moon or other celestial body** r = (140 + 1737) km = 1.877e6 m M = 7.348e22 kg G = 6.673e-11 T = 2*pi*sqrt( (1.877e6))/(6.673e-11*7.348e22) ) T = 7296 s = 2.02 h
Suppose that a planet was reported to be orbiting the sun-like star Iota Horologii with a period of 360 days. Find the radius of the planet's orbit, assuming that Iota Horologii has five times the mass as the Sun. (This planet is presumably similar to Jupiter, but it may have large, rocky moons that enjoy a pleasant climate. Use 2.00 1030 kg for the mass of the Sun.) 310 days x (24 hours / 1 day) x (3600 s/ 1 hr) = 26,784,000 s. 26,800,000 = 2*pi*√(a³/(6.67e-11 * 4.00e30)) 26,800,000 = 6.28*√(a³/2.67e20) 4,270,000 = √(a³/2.67e20) 1.82e13 = a³/2.67e20 a³ = 4.86e33 a = 1.69e11 m
A typical GPS (Global Positioning System) satellite orbits at an altitude of 5*10^7 m. (Astronomical data needed for this problem can be found on the inside back cover of the text.)(a) Find the orbital period of such a satellite. (Do B first) (b) Find the orbital speed of such a satellite. Earth's radius R = 6370 km = 6.37 * 10^6 m Earth's mass M = 6 * 10^24 kg v = sqrt(GM/r) = sqrt[6.67 * 10^-11 * 6 * 10^24/(5.637 * 10^7)] ---> (r = 6.37e6 + 5e7) = sqrt(7.1 * 10^6) Or v = 2.66 * 10^3 m/s = 2.66 km/s(a) Find the orbital period of such a satellite. 2*pi*r/v = 2 * 3.14 * 5.637 * 10^7/(2.66*10^3) = 133084 s = 133084/3600 hours = 36.967777 hours
Jill of the Jungle swings on a vine 7.3 m long. What is the tension in the vine if Jill, whose mass is 61 kg, is moving at 2.3 m/s when the vine is vertical? T - m*g = m*a = m*v^2/r so T = m*(g + v^2/r) = 61*(9.8 + 2.3^2/7.3) = 642N
A set of fossilized triceratops footprints show that the front and rear feet were d = 3.5 m apart, as shown in the figure below. The rear footprints were observed to be twice as deep as the front footprints. Assuming that the rear feet pressed down on the ground with twice the force exerted by the front feet, find the horizontal distance from the rear feet to the triceratops's center of mass. Divide distance by 3 3.5/3 = 1.1666m
In one hand you hold a 0.14 kg apple, in the other hand a 0.22 kg orange. The apple and orange are separated by 0.55 m. (a) What is the magnitude of the force of gravity that the orange exerts on the apple? GMm/r^2 (6.67e-11 * .14 * .22)/(.55^2) 3.738e-12 N (b) What is the magnitude of the force of gravity that the apple exerts on the orange? GMm/r^2 (6.67e-11 * .14 * .22)/(.55^2) 3.738e-12 N
Several volcanoes have been observed erupting on the surface of a planet's closest moon. Suppose that material ejected from one of these volcanoes reaches a height of 4.20 km after being projected straight upward with an initial speed of 240 m/s. (a) If the radius of this moon is 3303 km, outline a strategy that allows you to calculate its mass. Use kinematics to find the gravity of the moon. Then use the gravity in the equation g=GMm/r^2 to find the mass. (b) Use your strategy to calculate the moon's mass. v^2 = vo^2 + 2ax 0 = 240^2 + 2a(4200) -57600 = 8400a a or g = 6.857m/s^2
g = GM/r^2 M = gr^2/G M = (6.857)(3303000^2)/(6.67e-11) M = 6.3395e23kg
Find the orbital speed of a satellite in a circular orbit 810 km above the surface of the Earth. v = √ (GM / r ) = √ { GM / (R + h) } h = 810 x10^3 m R = 6.4 x10^6 m M = 6 x10^24 kg G = 6.67 x10^-11 N m^2/kg^2 sqrt((6.67e-11*6e24)/(810000+6.4e6)) V = 7.450 km/s
A 1350 kg car drives along a city street at 34.0 miles per hour, and a baseball has a mass of 0.142 kg. What speed must the baseball have if its momentum is to be equal in magnitude to that of the car? Give your result in miles per hour. m1 * V1 = m2 * V2 V2 = V1 * m1/m2 34*1350/0.142 = 323240 mi/hr
A 20.0 kg dog is running northward at 2.50 m/s, while a 5.00 kg cat is running eastward at 3.00 m/s. Their 70 kg owner has the same momentum as the two pets taken together. Find the direction and magnitude of the owner's velocity. Resultant momentum of the two √{ [ 20 *2.5]^2 + [5*3]^2 } = 52.20 kg m /s Owner's velocity = 52.2 / 70 = 0.75 m/s. -------------------------------------- tan θ = 50 / 15 θ = 73.3° (north of east)
Two air track carts move toward one another on an air track. Cart 1 has a mass of 0.39 kg and a speed of 1.2 m/s. Cart 2 has a mass of 0.63 kg. (a) What speed must cart 2 have if the total momentum of the system is to be zero? M1V1 = M2V2 0.39 * 1.2 = 0.63*V2 V2 = 0.39 * 1.2/0.63 V2 = 0.74 m/s (b) Since the momentum of the system is zero, does it follow that the kinetic energy of the system is also zero? No (c) Verify your answer to part (b) by calculating the system's kinetic energy. KE of cart 1 = (1/2)(0.39)(1.2^2) = 0.28 J KE of cart 2 = (1/2)(0.63)(0.74^2) = 0.17 J .28+.17 = .45J
A 220 g ball falls vertically downward, hitting the floor with a speed of 2.9 m/s and rebounding upward with a speed of 2.0 m/s. (a) Find the magnitude of the change in the ball's momentum. (Let up be in the positive direction.) Change in momentum = m(v-u) = 0.220(2.0 - (-2.9)) = 1.078 kg·m/s (b) Find the change in the magnitude of the ball's momentum. (Let negative values indicate a decrease in magnitude.) Magnitude of Momentum before = mu = 0.220*2.9 = 0.638 Ns Magnitude of Momentum after = mv = 0.220*2.0 = 0.440 Ns Change = 0.440 - 0.638 = -0.198 kg·m/s (c) Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor? The magnitude of the change in the ball's momentum. Explain. The answer in a is more directly related because momentum is a vector and any change must include the direction change
Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1415 N. Assume that the player's foot is in contact with the ball for 5.89 ^ 10-3 s. 1415 * 5.89*10^-3 = 8.33 kg·m/s
In a typical golf swing, the club is in contact with the ball for about 0.0014 s. If the 45 g ball acquires a speed of 62 m/s, estimate the magnitude of the force exerted by the club on the ball. 45 *10^-3* 62/0.0014 F = 1993 N = 1.993 KN
A 0.50 kg croquet ball is initially at rest on the grass. When the ball is struck by a mallet, the average force exerted on it is 185 N. If the ball's speed after being struck is 3.3 m/s, how long was the mallet in contact with the ball? a = f/m = 185/0.5 m/s^2 = 370 m/s^2 3.3 m/s / (370 m/s^2) = 0.0089189 sec 8.92 ms
A 15.0 g marble is dropped from rest onto the floor 1.44 m below. (a) If the marble bounces straight upward to a height of 0.640 m, what is the magnitude and direction of the impulse delivered to the marble by the floor? (Use a positive sign for the up direction, negative for down.) Impulse = change in momentum momentum of marble just prior to collison with floor: m1v1 = (0.015)(v1) (vertically downward is neg direction) v1 = √2gh (where h = 1.44) v1 = √19.6(1.44) v1 = √28.2 v1 = 5.31 m/s m1v1 = (0.015)(5.31) = - 0.08 kg-m/s (Momentum Before Floor)
Change in Momentum is: m2v2 - m1v1 = 0.053 - (- 0.08) = 0.133 kg-m/s (upward direction) Impulse = 0.133 kg·m/s (b) If the marble had bounced to a greater height, would the impulse delivered to it have been greater or less than the impulse found in part (a)? greater Explain. If the marble had bounced to a greater height then v2 would have been greater than 3.54 m/s and the Momentum After Floor would have become larger. Since the change in momentum is the difference between the two momentums with one momenutm being negative by its direction. The result is that their values are summed and so increasing one would increase the sum. The impulse is equal to the change in momentum so it too would be increased. Impulse would be greater than impulse in part (a).
To make a bounce pass, a player throws a 0.70 kg basketball toward the floor. The ball hits the floor with a speed of 5.7 m/s at an angle of 68° from the vertical. If the ball rebounds with the same speed and angle, what was the impulse delivered to it by the floor? 5.7m/s * sin(90-68) =5.7m/s * sin(22) = 2.14 2.14 * 2 =4.27 4.27 * .7 = 2.989 kg·m/s
A car with a mass of 950 kg and an initial speed of v1 = 21.0 m/s approaches an intersection, as shown in the figure. A 1300 kg minivan traveling northward is heading for the same intersection. The car and minivan collide and stick together. If the direction of the wreckage after the collision is 47.0° above the x axis. ** START WITH FINAL SPEED THEN SPEED OF MINIVAN Find the initial speed of the minivan. m2v2i = (m1+m2)Vsin37 (1300)v2i = (950+1300)(11.10sin37) v2i=19.22 m/s Find the final speed of the wreckage. m1v1i = (m1+m2)Vcos37 (950*21) = (m1+m2)Vcos37 19950 = (950+1300)Vcos37 Vcos37= 8.8667 V=11.10 m/s
A 0.440 kg block of wood hangs from the ceiling by a string, and a 0.072 kg wad of putty is thrown straight upward, striking the bottom of the block with a speed of 5.74 m/s. The wad of putty sticks to the block. (a) Is the mechanical energy of this system conserved?NO (b) How high does the putty-block system rise above the original position of the block? (0.072kg)(5.74m/s) = (0.072kg + 0.440kg)v v = 0.807m/s
Now from conservation of energy:
0.5(0.512kg)(0.807m/s)^2 = (0.512kg)(9.81m/s/s)h h = 0.0332m (or 3.32cm)
A 1200 kg car moving at 2.3 m/s is struck in the rear by a 2600 kg truck moving at 5.8 m/s.(a) If the vehicles stick together after the collision, is the final kinetic energy of the car and truck together greater than, less than, or equal to the sum of the initial kinetic energies of the car and truck separately? less Explain. 2600 kg impinges the less massive car, and with reference to the less massive car. the latter should be moving fast than the impinging car . Here both are moving with same velocity and it implies that there is loss of energy due to collision . (b) Verify your answer to part (a) by calculating the initial and final kinetic energies of the system. Velocity after collision V = (m1 u1 + m2 u2) / (m1 + m2) V = (1200*2.3 + 2600*5.8) / (3800) V = 4.7 m/s ================================ Initial kinetic energy = 0.5*(m1 u1 ²+ m2 u2²) = 0.5*((1200*2.3² + 2600*5.8) ²) = 46906 J Final kinetic energy 0.5(m1 + m2) V² = 3800*4.7² = 41971 J
14.5A 750 kg car stopped at an intersection is rear-ended by a 1730 kg truck moving with a speed of 14.5 m/s. If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of both vehicles after the collision. V2 = 2 m1 u1 / ( m1+m2) = 2 x 1730 x 14.5 / 2480 = 20.22 m/s (car) V1 = (m1-m2) u1 / (m1+m2) =(1730-750) 14.5 / 2480 = 5.73 m/s (truck)
The collision between a hammer and a nail can be considered to be approximately elastic. Estimate the kinetic energy acquired by a 11 g nail when it is struck by a 550 g hammer moving with an initial speed of 4.0 m/s. [2*mb/(ma + mb)]*vi (2*0.55)/(0.55+0.011) * 4.0m/s = 7.84m/s So K = 1/2*m*v^2 = 1/2*0.011*7.84^2 = 0.338 J
Two astronauts on opposite ends of a space ship are comparing lunches. One has an apple, the other has an orange. They decide to trade. Astronaut 1 tosses the 0.160 kg apple toward astronaut 2 with a speed of 1.22 m/s. The 0.130 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.10 m/s. Suppose that after the collision the orange moves exactly in the negative y direction with a speed of 1.23 m/s (not at the angle drawn in the figure). What is the final speed and direction of the apple in this case? So ma*vay = mo*voy (a - apple; o - orange) vay = mo*moy/ma = 0.130*1.23/.160 = 0.999m/s ( + y) So vax = (.160*1.22 -.130*1.10)/.160 = 0.326m/s So va = sqrt(.999^2 + .326^2) = 1.05m/s theta = arctan(.999/.326) = 71.9
In a nuclear reactor, neutrons released by nuclear fission must be slowed down before they can trigger additional reactions in other nuclei. To see what sort of material is most effective in slowing (or moderating) a neutron, calculate the ratio of a neutron's final kinetic energy to its initial kinetic energy, Kf / Ki for a head-on, elastic collision with each of the following stationary target particles. (Note: The mass of a neutron is m = 1.009 u, where the atomic mass unit, u, is defined as follows: 1 u = 1.66 10-27 kg.)(a) An electron (M = 5.49 10-4 u). FORMULA: [ (mass of neutron - mass of what you're trying to find)/(mass of neutron+mass of what you're trying to find) ] ^ 2 (b) A proton (M = 1.007 u). ((1.009-1.007)/(1.009+1.007))^2 = 9.84x10^-7 (c) The nucleus of a lead atom (M = 207.2 u). ((1.009-207.2)/(1.009+207.2))^2 = 0.981
A 72.5 kg tourist climbs the stairs of the Washington Monument, and stops at a height of 555 ft. How far does the Earth move in the opposite direction as the tourist climbs? First of all convert units to same standard (imperial or metric) Person will weigh 159.835lbs, planet Earth (check wiki) will be 1.317e25 lb d=159.835*555/1.317E25= 6.73E-24ft
A car moving with an initial speed v collides with a second stationary car that is 55% as massive. After the collision the first car moves in the same direction as before with a speed 31.5.(a) Find the final speed of the second car. MaUa= MaVa +MbVb Ma= Mb= 0.55Ma Va= 0.315Ua MaUa= Ma(0.315Ua) + 0.55MaVb MaUa- 0.315MaUa= 0.55MaVb (0.685MaUa)/.55Ma=Vb Vb=1.245 * v(b) Is this collision elastic or inelastic?
A 1.05 kg block of wood sits at the edge of a table, 0.760 m above the floor. A 0.0105 kg bullet moving horizontally with a speed of 725 m/s embeds itself within the block. What horizontal distance does the block cover before hitting the ground? (0.0105 x 725)/(1.05 + 0.0105) = 7.18m/sec. horizontal velocity. Time to drop vertically 0.760m = sqrt. (2h/g), = 0.155 secs. 7.18 x 0.155) = 1.1129m
During a severe storm in Palm Beach, Fla., in January 1999, 32 inches of rain fell in a period of nine hours. Assuming that the raindrops hit the ground with a speed of 12 m/s, estimate the average upward force exerted by one square meter of ground to stop the falling raindrops during the storm. (Note: One cubic meter of water has a mass of 1000 kg.) density = 1000 kg/m^3 volume = 32 inches CONVERT TO METERS= 0.8128 m^3 time = 9 hours = 32400 seconds velocity = 12 m/s 1000*.8128*12/32400 = 0.301 N
An apple that weighs 2.3 N falls vertically downward from rest for 1.6 s. (a) What is the change in the apple's momentum per second? 2.3M * 1s = 2.3 kg·m/s/s (b) What is the total change in its momentum during the 1.2 second fall? 2.3Nx1.6s=3.68 kg·m/s
To balance a 39.4 kg automobile tire and wheel, a mechanic must place a 60.2 g lead weight 25.0 cm from the center of the wheel. When the wheel is balanced, its center of mass is exactly at the center of the wheel. How far from the center of the wheel was its center of mass before the lead weight was added? 1.505Kg = 39.4Kg x distance 1.505Kg/39.4Kg = distance .0382 cm
A 74-kg canoeist stands in the middle of her 18-kg canoe. The canoe is 3.0 m long, and the end that is closest to land is 2.5 m from the shore. The canoeist now walks toward the shore until she comes to the end of the canoe. (a) When the canoeist stops at the end of her canoe, is her distance from the shore equal to, greater than, or less than 2.5 m? GREATER Explain. The center of gravity of the system will not move (ignoring water resistance). It remains 2.5 + 3/2 = 4 m from shore The canoe, being lighter, moves back more than the canoeist moves forward. (b) Verify your answer to part (a) by calculating the distance from the canoeist to shore. (74 + 18) 4 = 74 x + 18 (x+1.5) 368 = 74 x + 18 x + 32.4 335.6 = 92 x x = 3.65 m
A 62.4 kg canoeist stands in the middle of her canoe. The canoe is 3.5 m long, and the end that is closest to land is 2.5 m from the shore. The canoeist now walks toward the shore until she comes to the end of the canoe. Suppose the canoeist is 3.4 m from shore when she reaches the end of her canoe. What is the canoe's mass? The canoeists initial distance from the shore = 2.5 + 3.0 / 2 = 4.0 m. When she reaches the end of the canoe, she will have moved 4.0 - 3.4 = 0.6 m closer to the shore The canoe will have moved by 1.5 - 0.6 = 0.9 m away from the shore. So 0.6 * 63.6 = 0.9 * M M = 0.6 * 63.6 / 0.9 = 42.4 kg