AP Physics B: Work/Energy/Power Homework Challenge Sheet 1.

A 285-kg load is lifted 22.0 m vertically with an acceleration a = 0.160 g by a single cable. Determine (a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the load, (d) the work done by gravity on the load, and (e) the final speed of the load assuming it started from rest.

2.

A cyclist intends to cycle up a 7.8º hill whose vertical height is 150 m. Assuming the mass of bicycle plus cyclist is 75 kg, (a) calculate how much work must be done against gravity. (b) If each complete revolution of the pedals moves the bike 5.1 m along its path, calculate the average force that must be exerted on the pedals tangent to their circular

3.

An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should break when the elevator is at a height h above the top of the spring, calculate the value that the spring stiffness constant k should have so that passengers undergo an acceleration of no more than 5.0 g when brought to rest. Let M be the total mass of the elevator and passengers.

4.

A 0.620-kg wood block is firmly attached to a very light horizontal spring (k = 180 N m ) as shown. It is noted that the block–spring system, when compressed 5.0 cm and released, stretches out 2.3 cm beyond the equilibrium position before stopping and turning back. What is the coefficient of kinetic friction between the block and the table?

5.

A 280-g wood block is firmly attached to a very light horizontal spring. The block can slide along a table where the coefficient of friction is 0.30. A force of 22 N compresses the spring 18 cm. If the spring is released from this position, how far beyond its equilibrium position will it stretch on its first swing?

6.

Early test flights for the space shuttle used a “glider” (mass of 980 kg including pilot) that was launched horizontally at 500 km h from a height of 3500 m. The glider eventually landed at a speed of 200 km h . (a) What would its landing speed have been in the absence of air resistance? (b) What was the average force of air resistance exerted on it if it came in at a constant glide of 10º to the Earth?

7.

What minimum horsepower must a motor have to be able to drag a 310-kg box along a level floor at a speed of 1.20 m s if the coefficient of friction is 0.45?

8.

A bicyclist coasts down a 7.0º hill at a steady speed of 5.0 m s . Assuming a total mass of 75 kg (bicycle plus rider), what must be the cyclist’s power output to climb the same hill at the same speed?

Solutions: 1. (a) From the free-body diagram for the load being lifted, write Newton’s 2nd law for the vertical direction, with up being positive.

∑ F =F

T

 FT

− mg =ma =0.160mg →

(

)

= FT 1.16= mg 1.16 ( 285 kg ) 9.80 m = s2 3.24 × 103 N

 mg

(b) The net work done on the load is found from the net force. = Wnet F= d cos 0o net

= mg ) d ( 0.160

(

0.160 ( 285 kg ) 9.80 m s 2

=

9.83 × 103 J

(c)

The work done by the cable on the load is

W= FT d cos= 0o cable

(d)

(1.160mg= )d

(

1.16 ( 285 kg ) 9.80 m s 2

) ( 22.0 m )

m) ) ( 22.0=

7.13 × 10 4 J

The work done by gravity on the load is

(

WG = mgd cos180o = − mgd = − ( 285 kg ) 9.80 m s 2

(e)

−6.14 × 10 J ) ( 22.0 m ) = 4

Use the work-energy theory to find the final speed, with an initial speed of 0.

Wnet = KE2 − KE1 = 12 mv22 − 12 mv12 → = v2

2. (a)

2Wnet m

+= v12

(

2 9.83 × 103 J 285 kg

)= +0

8.31m s

The work done against gravity is the change in PE.

(

)

Wagainst = ∆PE = mg ( y2 − y1 ) = 1.1 × 105 J ( 75 kg ) 9.8 m s 2 (150 m ) = gravity

(b) The work done by the force on the pedals in one revolution is equal to the tangential force times the circumference of the circular path of the pedals. That work is also equal to the energy change of the bicycle during that revolution. Note that a vertical rise on the incline is related to the distance along the incline by rise = distance ( sinθ ) .

∆PE1 rev =∆ Wpedal = Ftan 2π r = mg ( y )1 rev = mgd1 rev sin θ → force

mgd1 rev sin θ = F= tan 2π r

( 75 kg ) ( 9.8 m s 2 ) ( 5.1 m ) sin 7.8o = 2π ( 0.18 m )

4.5 × 10 2 N

3. The maximum acceleration of 5.0 g occurs where the force is at a maximum. The maximum force occurs at the bottom of the motion, where the spring is at its maximum compression. Write Newton’s 2nd law for the elevator at the bottom of the motion, with up as the positive direction. Fnet = Fspring − Mg = Ma = 5.0 Mg

→

Fspring = 6.0 Mg

 mg

 Fspring

Now consider the diagram for the elevator at various points in its motion. If there are no non-conservative forces, then mechanical energy is conserved. Subscript 1 represents the elevator at the start of its fall, and subscript 2 represents the elevator at the bottom of its fall. The bottom of the fall is the zero location for gravitational PE ( y = 0 ) . There is also a point at the

Start of fall h

Contact with spring,

x Bottom of fall, 0 for gravitational PE

top of the spring that we will define as the zero location for elastic PE (x = 0). We have v1 = 0 ,

y1= x + h , x1 =0 , v2 = 0 , y2 = 0 , and x2 = x . Apply conservation of energy. E1 =E2 →

1 2

Mv12 + Mgy1 + 12 kx12 = 12 Mv22 + Mgy2 + 12 kx22 →

0 + Mg ( x + h ) + 0 = 0 + 0 + 12 kx22 → Mg ( x + h ) = 12 kx22 Fspring = 6.0 Mg = kx → x =

6.0 Mg k

 6 Mg  + h =  k 

→ Mg 

1 2

 6 Mg    k 

k

2

→

k=

12 Mg h

4. Use conservation of energy, including the non-conservative frictional force. The block is on a level surface, so there is no gravitational PE change to consider. The frictional force is given by = Ffr µ= F µ k mg , since the normal force is equal to the weight. Subscript 1 represents the k N

block at the compressed location, and subscript 2 represents the block at the maximum stretched position. The location of the block when the spring is neither stretched nor compressed is the zero location for elastic PE (x = 0). Take right to be the positive direction. We have v1 = 0 , x1 = −0.050 m , v2 = 0 , and x2 = 0.023 m . WNC + E1 = E2 → Ffr ∆x cos180o + 12 mv12 + 12 kx12 = − µ k mg ∆x + 12= kx12

(

1 2

1 2

mv22 + 12 kx22 →

kx22 →

)  ) (180 N m ) ( 0.050m ) − ( 0.023m = 2 ( 0.620 kg ) ( 9.80 m s ) ( 0.073 m )

k x12 − x22 µk = = 2mg ∆x

2

2

2

0.40

5. Use conservation of energy, including the non-conservative frictional force. The block is on a level surface, so there is no gravitational PE change to consider. Since the normal force is equal to the weight, the frictional force is= Ffr µ= F µ k mg . Subscript 1 represents the block at the k N compressed location, and subscript 2 represents the block at the maximum stretched position. The location of the block when the spring is neither stretched nor compressed is the zero location for elastic PE (x = 0). Take right to be the positive direction. We have v1 = 0 , x1 = −0.18 m , and v2 = 0 . The value of the spring constant is found from the fact that a 22-N

force compresses the spring 18 cm, and so= k F= x 22 N 0.18= m 122.2 N m . The value of x2 must be positive.

WNC + E1 = E2 → Ffr ∆x cos180o + 12 mv12 + 12 kx12 =

mv22 + 12 kx22 →

 2 µ k mg  x1 + x12  = 0 → k  k  2 ( 0.30 )( 0.28 )( 9.80 )  2 ( 0.30 )( 0.28 )( 9.80 ) 2  x22 + x2 −  ( −0.18 ) + ( −0.18 ) = 0 → 122.2 122.2  

− µ k mg ( x2 − x1 ) + 12 kx12 = 12 kx22 → x22 +

2 µ k mg

1 2

x2 − 

x22 + 0.01347 x2 − 0.02997 = 0 → x2 = 0.1665 m, − 0.1800m → x2 = 0.17 m 6. (a) If there is no air resistance, then conservation of mechanical energy can be used. Subscript 1 represents the glider when at launch, at subscript 2 represents the glider at landing. The landing location is the zero location for elastic PE (x = 0). We have y1 = 500 m , y2 = 0 , and

 1m s  = v1 500 = km h   138.9 m s . Solve for v2  3.6 km h  E1 =E2 →

1 2

mv12 + mgy1 = 12 mv22 + mgy2 →

(

v2 = v12 + 2 gy1 = (138.9 m s ) + 2 9.80 m s 2 2

) ( 3500 m ) =296 m s  3.61mkms h  



= 1067 km h ≈ 1.1 × 10 km h 3

(b) Now include the work done by the non-conservative frictional force. Consider the diagram of the glider. Calculate the work done by friction.

WNC = Ffr d cos180o = − Ffr d = − Ffr

3500 m

sin10o Use the same subscript representations as above, with y1 , v1 , and y2 as before, and

 1m s  = v2 200 = km h   55.56 m s . Write the energy conservation equation and solve for  3.6 km h  the frictional force. 7. Draw a free-body diagram for the box being dragged along the floor. The box has a constant speed, so the acceleration is 0 in all directions. Write Newton’s 2nd law for both the x (horizontal) and y (vertical) directions.

∑F ∑F

y

= FN − mg = 0 → FN = mg

x

= FP − Ffr = 0 → FP = Ffr = µ k FN = µ k mg

The work done by FP in moving the crate a distance ∆x is given by W = FP ∆x cos 0o = µ k mg ∆x . The power required is the work done per unit time.

W µ k mg ∆x ∆x = = µ k mg = µ k mgv= x t t t  1 hp  1641W   = 2.2 hp  746 W  P =

( 0.45 )( 310 kg ) ( 9.80 m

)

s 2 (1.20 m s= ) 1641W

8. First, consider a free-body diagram for the cyclist going down hill. Write Newton’s 2nd law for the x direction, with an acceleration of 0 since the cyclist has a constant speed.

∑F

= mg sin θ − Ffr = 0 → Ffr = mg sin θ

x

Now consider the diagram for the cyclist going up the hill. Again, write Newton’s 2nd law for the x direction, with an acceleration of 0.

∑F

x

= Ffr − FP + mg sin θ = 0 → FP = Ffr + mg sin θ

Assume that the friction force is the same when the speed is the same, so the friction force when going uphill is the same magnitude as when going downhill. FP = Ffr + mg sin θ = 2mg sin θ

The power output due to this force is given by Eq. 6-17.

(

= P F= v 2mgv sin= θ 2 ( 75 kg ) 9.8 m s 2 P =

9.0 × 10 2 W

) ( 5.0 m s ) sin 7.0

o