AP Physics. Chapters 7 & 8 Review

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1 AP Physics Chapters 7 & 8 Review

2 1.A particle moves along the x axis and is acted upon by a single conservative force given by F x = ( x)N where x is in meters. The potential energy associated with this force has the value +30J at the origin (x = 0). What is the value of the potential energy at x = 4.0 m? ΔK = W = F( x)dx ΔK = 4 m ( x)dx = 20x 2x 2 0 ΔK = ( 80 32)J = 48J ( ) 0 4 m E Total = U + K +30J = U + 48J U = 18J 2

3 2.The only force acting on a 0.50-kg object as it moves along the x axis is given by F x = ( 4x 3 )N, where x is measured in m meters. If the speed of the object at x = 0 is equal to 4.0, what is its speed at x = 2.0 m? ΔK = W = F x ( )dx ΔK = ( 4 x 3 )dx = x 4 0 2m ΔK =16J ( ) 0 2m ΔK = 1 2 mv 2 f 1 2 mv i2 16J = 1 2 mv 2 f 4J 20J = 1 2 mv 2 f 2ΔK m = v f = v f = 8.94 m 40J 0.5kg

4 3. A 0.40-kg particle moves under the influence of a single conservative force. At point A where the particle has a speed of 10 m, the potential energy associated with the conservative force is +40J. As the particle moves from A to B, the force does +25J of work on the particle. What is the value of the potential energy at point B? at point B, W = ΔK if ΔK = 25J, then K = ( )J = 45J U = 40J 25J = 15J at point A U = +40J K = 1 2 mv 2 = 1 2 K = +20J E T = U + K 0.4Kg ( ) 10 m E T = 40J + 20J = 60J because U + K = 60J ( ) 2 4

5 k = 800 N m 12Kg P #4 W Net = ΔK W W = 1 P S 2 mv 2 1 f 2 mv 2 o What is the speed of the block when it is 13 cm from its equilibrium position? F x 1 2 kx 2 = 1 2 mv 2 2( F x) kx 2 m = v 2( 80N 0.13m) ( 800 N )( 0.13m) 2 m 12Kg 5 = v v = 0.78 m

6 5. What maximum angle does the string make with the vertical as the object swings? U Top = K B mgh = 1 mv 2 2 B h = v 2 B 2g = ( 4 m ) 2 = 0.816m 19.6 m 2 L Lcosθ = h L h = L cosθ L h L = cosθ cos 1 L h = θ L 2m 0.816m cos 1 = θ 2m 53.7 = θ 6

7 6. A 2.0kg mass swings at the end of a light string of length 3.0m. Its speed at the lowest point on its circular path is m 6.0. What is its kinetic energy at an instant when the string makes an angle of 50 with the vertical? K max = 1 mv 2 2 max = ( 0.5) ( 2kg) 6 m ( ) 2 = 36J h = L Lcosθ = L 1 cosθ ( ) h = 3m ( ) 1 cos50 ( ) =1.07m U = mgh = ( 2kg) ( 9.8 N )( 1.07m) kg U = 21J E Total = K max K max = K + U K = K max U K = 36J 21J K =15J 7

8 7.What is the value of the kinetic energy of the ball at B minus the kinetic energy of the ball at A? A 10m E Top = E Bottom 30m U A + K A = K B B U A = K B K A U A = mgh U = ( 0.04Kg) ( 9.8 N )( 30m) A Kg U A =11.76J = K B K A 8

9 8. A 1.2-kg mass is projected from ground level with a velocity m of 30 at some unknown angle above the horizontal. A short time after being projected, the mass barely clears a 16 meter tall fence. Disregard air resistance and assume the ground is level. What is the kinetic energy of the mass as it clears the fence? E i = E f K i = U F + K F K i U F = K F 1 2 mv i2 mgh = K F ( 0.5) ( 1.2kg) ( 30 m ) 2 ( 1.2kg) ( 9.8 N )( 16m) = K kg F 540J J = K F K F = 352J 9

10 R Rcosθ E A = E B K A + U A = K B 1 mv 2 2 A + mgh = 1 mv 2 2 B v A 2 + 2gh = v B 2 v A 2 + 2gh = v B h = R Rcosθ ( ) = 7.02m 12 m ( ) m v B = 16.8 m ( ) 7.02m 2 ( ) = v B 9. A skier weighing 0.80kN comes down a frictionless ski run that is circular (R = 30 m) at the bottom, as shown. If her speed is 12 m at point A, what is her speed at the bottom of the hill (point B)? 10

11 10. A spring (k = 800 N m ) is placed in a vertical position with its lower end supported by a horizontal surface. The upper end is depressed 20 cm, and a 0.15Kg block is placed on the depressed spring. The system is then released from rest. How far above the point of release will the block rise? E B = E Top U Spring U block = U Top 1 2 kx 2 mgh = mgh ( ) 0.2m 800 N m Kg ( ) 2 ( ) 9.8 m ( ) 0.2m = H m = h 11

12 11. A 4.0-kg block, sliding on a horizontal frictionless surface, is N attached to one end of a horizontal spring ( k =100 m ) which has its other end fixed. If the maximum distance the block slides from the equilibrium position is equal to 20 cm, what is the speed of the block at an instant when it is a distance of 16 cm from the equilibrium position? U S = U 16cm + K 16cm 1 2 ka2 = 1 kx mv ka 2 = kx 2 + mv 2 k( A 2 x 2 ) = mv 2 v = 100 N m ( 4kg) v = 0.6 m (( 0.2m) 2 ( 0.16m) 2 ) v = k m ( A 2 x 2 ) 12

13 k = 400 N m 8kg 12. The blocks shown are released from rest with the spring unstretched. The pulley and the horizontal surface are frictionless. What is the maximum extension of the spring? 4kg h = x 13 E G = E S U G = U S mgh = 1 2 kx 2 mgx = 1 2 kx 2 2mg = kx 2mg k ( ) 2 4.0kg ( ) 9.8 N Kg = x = x 400 N m 0.196m = x 19.6cm = x

14 13. A 1.0-kg block is released from rest at the top of a frictionless incline that makes an angle of 37 with the horizontal. An unknown distance down the incline from the point of release, there is a spring with k = 200 N. It is observed m that the mass is brought momentarily to rest after compressing the spring 30cm. What distance does the mass slide from the point of release until it is brought momentarily to rest? U Top = U Spring at the Bottom mgh = 1 2 kx 2 mg( dsinθ) = 1 kx 2 2 mg( d 3 ) = 1 5 kx 2 2 d = 5kx 2 6mg ( ) 0.3m d = N m 6 1Kg d =1.53m ( ) 2 ( ) ( ) 9.8 N Kg 14

15 14. The same constant force is used to accelerate two carts of the same mass on frictionless tracks. The force is applied to cart A twice as long as it is applied to cart B. The work the force does on A is W A ; that on B is W B. Which statement is correct? F = ma a = Δv Δt aδt = Δv a 2Δt ( ) = 2Δv If F is constant, then a is constant. W = ΔK = 1 m ( v ) 2 2 4W = 1 m ( 2v ) 2 2 W A = 4W B 15

16 15. A 10-kg object is dropped from rest. After falling a distance of 50 m, it has a m speed of 25. How much work is done by the dissipative (air) resistive force on the object during this descent? E Top = E Bottom U Top W f = K Bottom U Top K Bottom = W f mgh 1 2 mv 2 = W B f ( 10Kg) ( 10 N )( 50m) 1 Kg 2 10Kg ( ) 25 m ( ) 2 = W f 5000J 3125J = W f 1875J = W f 16

17 16. A 1.2-kg mass is projected down a rough circular track of radius 2.0 m as shown. The speed of the mass at point A is m 3.2, m and at point B, it is 6.0. How much work is done on the mass between A and B by the force of friction? A B R K A + U A W f = K B K A + U A K B = W f 1 mv 2 2 A + mgr 1 mv 2 2 B = W f 6.144J J 21.6J = W f 8.06J = W f 17

18 17. A 25-kg block on a horizontal surface is attached to a light spring of k = 8.0 kn m. The block is pulled 10 cm to the right from its equilibrium position and released from rest. When the block has moved 2.0 cm toward its equilibrium position, its kinetic energy is 12 J. How much work is done by the frictional force on the block as it moves the 2.0 cm? U 10cm W f = U 8cm + K U 10cm U 8cm K = W f 1 2 ka2 1 2 kx 2 K = W f 1 k ( A 2 2 x 2 ) K = W 8000 N m f (( ) 2 ( 0.08m) 2 ) 12J = W f 0.1m J 12J = W f = 2.4J 18

19 2Kg U Top + K Top W f = E B = 0 2J = 1 2 v = d 20 ( 2Kg )v 2 2 m U Top + K Top = W f mgh mv 2 = µ ( mgcosθ)d gdsinθ v 2 = µ ( gcosθ)d 20. A 2.0-kg block is projected down a plane that makes an angle of 20 with the horizontal with an initial kinetic energy of 2.0 J. If the coefficient of kinetic friction between the block and plane is 0.40, how far will the block slide down the plane before coming to rest? 19

20 #20.2 gd sinθ v 2 = µ ( gcosθ)d 1 m m 2 2 = µ ( gcosθ)d gd sinθ = g ( µcosθ sinθ)d 1 m 2 2 ( ) = d g µcosθ sinθ 1 m 2 2 ( ) = d = 3m 9.8 m

21 21. Two equal masses are raised at constant velocity by ropes that run over pulleys, as shown below. Mass B is raised twice as fast as mass A. The magnitudes of the forces are FA and FB, while the power supplied is respectively PA and PB. Which statement is correct? Since v is constant, F B = F A and W B = W A 21

22 P = W t F B = F A P B = W B 1 t P A = W A t 2 P B = 2 W B t = 2 W A t P B = 2P A The answer is b. 22

= 1 2 kx2 dw =! F! d! r = Fdr cosθ. T.E. initial. = T.E. Final. = P.E. final. + K.E. initial. + P.E. initial. K.E. initial =

= 1 2 kx2 dw =! F! d! r = Fdr cosθ. T.E. initial. = T.E. Final. = P.E. final. + K.E. initial. + P.E. initial. K.E. initial = Practice Template K.E. = 1 2 mv2 P.E. height = mgh P.E. spring = 1 2 kx2 dw =! F! d! r = Fdr cosθ Energy Conservation T.E. initial = T.E. Final (1) Isolated system P.E. initial (2) Energy added E added

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