Physics 110 Spring 2006 Work and Energy Problems – Their Solutions 1. A block slides down a curved frictionless tract and then up an incline. The coefficient of kinetic friction between the block and the incline is µk. Using energy methods show that the maximum height reached by the block h is ymax = . 1 + µ k cot θ

Using conservation of energy, the kinetic energy at the bottom of the ramp is due to the change in potential energy as the block falls through a height h. The kinetic energy at the bottom of the ramp is transferred to potential energy as the block rises up the ramp to a height ymax and some is lost to friction. The energy loss due to friction is given by the work done, which is the product of the frictional force and the distance the block travels up the ramp. The distance the block travels up y y the ramp is given by sin θ = max → d = max and the force due to friction is sin θ d given by µ k FN = µ k mg cos θ . Therefore we have using conservation of energy y mgh = mgymax + µ k mg cos θ × max = (mg + µ k mg cot θ ) ymax sin θ h ∴ h = (1 + µ k cot θ ) ymax → ymax = (1 + µ k cot θ ) 2. Suppose that the Nott Memorial is topped with an approximately hemispherical dome. Suppose that the dome is frictionless when wet. Somehow an individual has balanced a pumpkin at the top to the dome at an angle of θi = 0o with the vertical. Suppose that on a rainy night, a gust of wind starts the pumpkin sliding from rest. It loses contact with the dome when the line from the center of the hemispherical dome to the pumpkin makes a certain angle with respect to the vertical. At what angle does this happen? mv 2 → v 2 = Rg cos θ , R where we have set FN = 0 when the pumpkin leaves the surface. In order to find the angle we need to use conservation of energy: mgR = 12 mv 2 + mgR cos θ . The initial energy is mgR (taking the dome level as the zero of the potential energy) From the free body diagram we have FN − mg cos θ = −

and the final energy when the pumpkin leaves the surface is ½ mv2 + mgh, where h is the height of the pumpkin above the base of the dome. Therefore we have using the speed of the pumpkin above, gR = 12 gR cos θ + gR cos θ = 32 gR cos θ → cos θ = 23 → θ = 48.2 o 3. A child of mass m rides on an irregularly curved slide of height h = 2m. The child starts from rest at the top of the slide. a. What is the speed of the child at the bottom of the slide assuming that no friction is present (the slide is wet)? b. If a force of kinetic friction acts on the child, how much mechanical energy does the system lose? Assume that vf = 3.0 m/s and that mchild = 20kg. a. Here the normal force does no work on the child since this force is always perpendicular to the displacement of the child. Since there is no friction, mechanical energy is conserved and taking the y-coordinate in the upward direction from the bottom of the slide. Thus, yi = h and yf = 0 and applying conservation of energy we have mgh = 12 mv 2 → v = 2 gh = 2 × 9.8 sm2 × 2m = 6.26 ms . b. When there is friction, mechanical energy is not conserved. The energy loss due to friction is given from Ei = E f + ∆E → ∆E = Ei − E f = mgh − 12 mv 2 . 2 ∆E = 20kg × 9.8 sm2 × 2m − 12 × 20kg × (3.0 ms ) = 392 J − 90 J = 302 J 4. A child slides without friction from a height h along a curved water slide. If she is launched from a height h/5 into the pool, what is her maximum airborne height y in terms of h and θ?

By conservation of energy, we have for the velocity of the girl when she leaves the

h 1 2 ⎛4 ⎞ + 2 mv → v = 2 g ⎜ h ⎟ , where the 5 ⎝5 ⎠ y-component of her velocity is vsinθ. Now she is a projectile in flight, so in the vertical direction, since her x-velocity is a constant using conservation of energy slide mgh = mg

mgy = mg

2 h 1 2 h vy h 1 ⎛ ⎛ 4 ⎞⎞ 2 h 4h + 2 mv y → y = + = + × ⎜⎜ 2 g ⎜ h ⎟ ⎟⎟ sin θ → y = + sin 2 θ 5 5 2g 5 2g ⎝ ⎝ 5 ⎠⎠ 5 5

5. A child starts from rest and slides down a frictionless slide. In terms of R and H, at what height h will he lose contact with the section of radius R?

Starting with conservation of energy applied between the top of the slide and when the child leaves the slide at a height h we have for the velocity of the child mgH = mgh + 12 mv 2 → v = 2 g (H − h ) . Next we use Newton’s laws to relate h and H. The maximum speed that the child can have before losing contact with the slide is when the normal force vanishes. Thus the component of the weight of the child gives rise to the centripetal force. Thus we v 2 2 gm(H − h ) have mg sin θ = m = . For small angles, we have that sinθ ~ h/R R R and using this we have a relation between h and H, h 2 gm(H − h ) namely mg = → h = 23 H R R 6. The drawing below shows a plane diving toward the ground and then climbing back upward. During each part of the motion, the lift force, L acts perpendicular to the displacement s, and has magnitude 1.7x103m in both cases. The thrust, T is provided by the engines and points in the direction of the displacement, and has the same magnitude during the dive and the climb. The weight of the plane, W has magnitude 5.9x104 N. In both cases the net work done is due to the combined action of the forces L, T, and W. a. Is more net work done during the dive or the climb? Explain b. Find the difference between the net work done during the diving and climbing motions.

a. In both cases, the lift force L is perpendicular to the displacement of the plane, and, therefore, does no work. As shown in the drawing when the plane is in the dive, there is a component of the weight W that points in the direction of the displacement of the plane. When the plane is climbing, there is a component of the weight that points opposite to the displacement of the plane. Thus, since the thrust T is the same for both cases, the net force in the direction of the displacement is greater for the case where the plane is diving. Since the displacement s is the same in both cases, more net work is done during the dive . b. The work done during the dive is Wdive = (T + W cos 75°) s , while the work done during the climb is Wclimb = (T + W cos 115°) s . Therefore, the difference between the net work done during the dive and the climb is Wdive – Wclimb = (T + W cos 75°) s – (T + W cos 115°) s = Ws (cos 75° – cos 115°) = (5.9 × 10 4 N)(1.7 × 103 m)(cos 75° – cos 115°) = 6.8 × 10 7 J 7. A 1200kg car is being driven up a 5o hill. The frictional force is directed opposite the motion of the car and has magnitude of Ffr = 524N. A force F is applied to the car by the road and propels the car forward. If the length of the road is 290m, what should the magnitude of F be so that the net work done by all of the forces acting on the car is +150kJ? The net work done on the car is WT = WF + Wf + Wg + WN WT = Fs cos 0.0° + Ffr s cos 180° – mgs sin 5.0° + FNs cos 90°. Rearranging this result gives WT F= + f + mg sin 5.0° s =

(

)

150 × 103 J 2 3 + 524 N + (1200 kg ) 9.80 m/s sin 5.0° = 2.07 × 10 N 290 m

8. A water skier lets go of the tow rope upon leaving the end of a jump ramp at a speed of 14m/s. The skier has a speed of 13m/s at the highest point of the jump. What is the skier’s height H above the top of the ramp if air resistance is ignored?

Conservation of energy gives

1 mv 2 f 2

+ mghf =

1 mv 2 0 2

+ mgh0

Rearranging gives hf − h0 =

(14.0 m/s )2 − (13.0 m/s )2 2 ( 9.80 m/s 2 )

= 1.4 m

9. A wrecking ball swings at the end of a 10.0m long cable on a vertical circular arc. The crane operator manages to give the ball a speed of 6.5 m/s as the ball passes through the lowest point of its swing and then gives the ball no further help. If friction and air resistance are negligible, what is the speed vf does the ball have then the cable makes and angle of 32o with respect to the vertical? Since energy is conserved, the ball will have the same speed at the bottom of its swing whether it is moving toward or away from the crane. As the cable swings to an angle of 32.0°, the ball will rise a distance h = L (1 – cos 32.0°), where L is the length of the cable. Conservation of energy then gives 12 mv02 = 12 mvf2 + mgh Solving for vf gives

vf = v02 − 2 gL (1 − cos 32.0° ) =

( 6.50 m/s )2 − 2 ( 9.80 m/s2 ) (10.0 m )(1 − cos 32.0°) =

3.53 m/s

10. A projectile of mass 0.75kg is launched straight up in the air with an initial speed of 18m/s. a. How high would the projectile go if there were no air friction? b. If the projectile rises to a maximum height of 11.8m, what is the average force exerted on the projectile due to air resistance? a. The conservation of mechanical energy states that 1 mv 2 + mgh = 1 mv 2 + mgh f 0 2 4 2 4 14 244 3f 14 244 30

Ef

E0

The mass m can be eliminated algebraically from this equation since it appears as a 1 2

factor in every term. Solving for the final height hf gives hf =

( v02 − vf2 ) + h

0 g Setting h0 = 0 m and vf = 0 m/s, the final height, in the absence of air resistance, is

hf =

vo2 − vf2 2g

=

(18.0 m / s )2 − ( 0 m/s )2

(

2 9.80 m / s

2

)

b. The work-energy theorem is Wnc =

(

= 16.5 m

1 mv 2 f 2

2

− 12 mv0

) + ( mgh

f

− mgh0 ) where Wnc

is the non-conservative work done by air resistance. According to Equation 6.1, the work can be written as Wnc = ( FR cos 180° ) s , where FR is the average force of air resistance. As the projectile moves upward, the force of air resistance is directed

downward, so the angle between the two vectors is θ = 180° and cos θ = –1. The magnitude s of the displacement is the difference between the final and initial heights, s = hf – h0 = 11.8 m. With these substitutions, the work-energy theorem

(

)

becomes − FR s = 12 m vf2 − vo2 + mg ( hf − h0 ) . Solving for FR gives FR =

=

1m 2

( vf2 − vo2 ) + mg ( hf − h0 ) −s

1 2

( 0.750 kg ) ⎡⎣( 0 m/s )2 − (18.0 m/s )2 ⎤⎦ + ( 0.750 kg ) ( 9.80 m/s2 ) (11.8 m ) = − (11.8 m )

2.9 N

11. At a carnival, you try to impress your significant other by trying to ring a bell at the top of a pole by striking a target with a 9.0kg hammer. In response, a 0.40kg metal piece is sent up the pole toward the bell, which is located 5.0m above the ground. Suppose that 25% of the hammer’s kinetic energy is used to do the work of sending the metal piece upward, how fast much the hammer be moving when it strikes the target so that the bell just barely rings? According to the work-energy theorem we have Wnc =

(

1 2

mvf2 + mghf

)−(

1 2

mv02 + mgh0

)

The metal piece starts at rest and is at rest just as it barely strikes the bell, so that vf = v0 = 0 m/s. In addition, hf = h and h0 = 0 m, while W = 0.25 1 M v 2 , nc

(

2

)

where M and v are the mass and speed of the hammer. Thus, the work-energy theorem becomes 0.25

(

1 2

Mv

2

) = mgh . Solving for the speed of the hammer, we

find

v =

2mgh = 0.25 M

2(0.400 kg)(9.80 m/s2 )(5.00 m) = 4.17 m/s 0.25 (9.00 kg)

12. A person is making homemade ice cream. She exerts a force of magnitude 22N on the free end of a crank handle, and this end moves in a circular path of radius 0.28m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 1.3s, what is the average power being expended by the woman? The work done by the person in turning the crank one revolution is W = F (2π r) = (22 N)(2π)(0.28 m) = 39 J. Then the power expended is W 39 J simply P = = = 3.0 × 101 W t 1.3 s