Energy Review Questions (click here for a downloadanble pdf of the anwsers)

 

1.   A net force of 90.0 N does 45.0 J of work on a brick.  What is the magnitude of the displacement of the brick?

Solution:

W = Fd
So d = W/F

d = 0.50 m

 

2.   How much work is done against gravity in lifting a 3.0 kg object through a distance of 40.0 cm ?

Solution:

W = F x d,      where F = mg

W = mg x d

W = 3.0 x 9.8 x 0.4

W = 12 J

 

3.   A ladder 3.0 m long and weighing 2.0 x 102 N has its center of gravity 1.20 m from the bottom.  At its top end is a 50.0 N weight.  Compute the work required to raise the ladder from a horizontal position on the ground to a vertical position.

Solution:

Total work = work lifting ladder + work lifting load

Wt = Wladder + Wload

Wt = Fdladder + Fdload

Wt = 200 x 1.2 + 50 x 3

Wt = 390 J

 

4.   Which of the following are units for kinetic energy?
     (I) kgm/s       (II) kgm/s2     ( III) N m      (IV) kgm2/s2      (V) W¥s

Solution:

III, IV, and V are units of energy

 

5.   Which one of the following in not an expression for energy?
   (I) mgh    (II) mad    (III) Pt     (IV) Fd     (V) ma 

Solution:

V is not an expression for energy

Ep = mgh

W = Fd, and F = ma, so W = mad

E = Pt

 

6.   Consider the four units given below.
        I. kgm/s2        II. J      III. N¥m           IV. kWh

    
Which of the above are units of energy?

Solution:

II, III, and IV are units of energy

 

7.   What is the kinetic energy of a 60 kg athlete running 6.0 m/s?

Solution:

Ek = 0.5 mv2

Ek = 0.5 x 60 x 62

Ek = 1080 J

 

8.   How would the kinetic energy of a moving object be affected if its velocity were doubled while the mass remains constant.

Solution:

Since Ek = 0.5 mv2 doubling the velocity will produce a four fold increase since velocity is squared.

Ek increases by 4 times

 

9.   How much energy would 1.0 kg of water falling 15 m generate?

Solution:

Ep = mgh
Ep = 1.0 x 9.8 x 15

Ep = 147 J

 

10. When an object is lifted 10 meters it gains a certain amount of potential energy.  If the same object is lifted 20 m, its potential energy is

Solution:

Since PE is proportional to h, if h increases by a factor of two so does the PE.

 

11. How much power is produced by a 60.0 kg person running up a 4.5 m high flight of stairs in
3.0 sec?

Solution:

P = W/t          where W = F x d        and F = mg

P = mgd / t

P = 60 x 9.8 x 4.5/ 3.0

P = 882 W

 

12. A 20 kg load of shingles is carried a vertical height of 8.0 m up a ladder in 40 s. (g = 10 N/kg) The power generated is approximately

Solution:

P = W/t
P = Fd/t
P= mgd/t    40 W

 

13. A machine with a power rating of 15.0 kW must complete a job requiring an expenditure of
1.5 x 105 J of energy.  How long will it take to complete the job?

Solution:

Since P = E/t
t = E/P

t = 15 000 / 1.5 x 105

t = 10.0 s

 

14. A 0.50 kg block is sliding across a table top with an initial velocity of 0.20 m/s.  It slides to rest in a distance of 0.70 m.  Find the average friction force that retarded its motion.

Solution:

ÆEk = W = Ff x d

Ff = ÆEk / d

Ff = 0.014 N

 

15. A motorcycle travelling at 45 km/h has 135,000 J of K.E.  If a force of 540 N acts on the motorcycle after the engine is turned off, how far does the motorcycle travel before it stops?

Solution:

W = ÆEk

d = ÆEk/ F

d = 135000 / 540

d = 250 m

 

16. A force of 1.50 N acts on a 0.20 kg cart so as to accelerate it along an air track.  How fast is the cart going after acceleration from rest through 30 cm if friction is negligible?

Solution:

W = ÆEk

Fd = 0.5 mv2

v = Ã(2Fd/m)

v = 2.1 m/s

 

17. A force of 35 N accelerates a 2.0 kg object from rest for a distance of 5.0 m along a level frictionless surface, the force then changes to 25 N and acts for and additional 3.0 m.

a) What is the final kinetic energy of the object?

b) How fast is it moving?

Solution:

For the initial force:  Fd = 35 x 5 = 165 J

For the second force Fd = 25 x 3 = 75 J

a) Total work = ÆEk = 240 J

b) v = Ã(2ÆEk/m)     v = 15.5 m/s

 

18. Conservation of energy means that

Solution:

Energy can be transformed from one form to another but the total amount remains the same

 

 

Energy Calculations 2

 

1.   A car of mass 1350 kg travelling at 35.0 m/s along a highway runs out of gas and starts to coast at the bottom of a hill 15.0 m high.  The length of the road is 540.0 m and the force of friction on the car is 750.0 N.  Will the car make it to the top of the hill?

i) if so, what will its speed be at that point?

ii) if not, how much energy would it need to reach the top?

Solution:

a)  Etotal before = Etotal after

Ek  = Ek' + Ep' + Eheat

0.5 mv2 = 0.5 mv'2 + mgh' + Eheat

0.5 mv'2 = 0.5 mv2 - mgh' - Eheat

0.5 mv'2 = 0.5 x 1350 x 352 - 1350 x 9.81 x 15.0 - 750 x 540

0.5 mv'2 = 2.2 x 105 J

v' = 18.2 m/s

b) not needed, since the car reaches the top

 

2.   A 7.50 kg bowling ball, which is smooth and therefore experiences an insignificant amount of air friction at low speeds is carried to the top of a 10.0 m driving tower and dropped.

a) how much G.P.E. does the ball gain?

b) relative to the bottom of the tower, what is the total energy of the ball when it is at:

      i) the top of the tower?

      ii) a point 5.3 m from the top?

c) what is the K.E. of the bowling ball when it is at :

      i) a point 5.3 m from the  top?

      ii) the bottom of the tower?

d) what is the velocity of the ball when it is at :

      i) a point 5.3 m from the top? 

      ii) the bottom of the tower?

Solution:

a) Ep = mgh = 7.5 x 9.81 x 10 = 736 J

b) i) Et = Ep + Ek = Ep + 0 = mgh = 7.5 x 9.81 x 10 = 736 J

b) ii)  Et = constant = 736 J

c) i) Ek = Et - Ep = 736 - 7.5 x 9.8 x 4.7 = 391 J

c) i) Ek = Et = 736 J

d) i) v = Ã(2Ek/m) = 10.2 m/s

d) i) v = Ã(2Ek/m) = 14.0 m/s

 

3.   A 5.0 kg stone is released 20.0 m above the ground.  What is its velocity when it hits the ground?

Solution:

Ep = Ek

mgh = 0.5 mv2

v = Ã(2gh)

v = Ã(2 x 9.81 x 20)

v = 19.8 m/s

 

4.   A 3.0 kg metal ball at rest is hit by a 1.0 kg metal ball moving at 4 m/s. The 3.0 kg ball moves forward at 2.0 m/s and the 1.0 kg ball bounces back as a result of the collision.

a) what is the total kinetic energy of the two balls before the collision?

b) what is the total K.E. of the two balls after the collision?  (assume a perfectly  elastic collision)

c) how much energy is transferred during the collision from the small ball to the large ball?

d) what is the velocity of the small ball after the collision?

Solution:

a) Ek = 0.5 m1v2 + 0.5 m2v2

    Ek = 8 J

b) Ek after = 8 J if the collision is perfectly elastic

c) Ek = 0.5mv'2 = 6J

d) ÆEk = 2 J

     v = Ã(2ÆEk/m) = 2m/s

 

5.   A car of mass 1700 kg coasting up a hill has a speed of 13.9 m/s at a point  15.7 m above the bottom of a hill.  By the time it has reached the top of the hill, a distance of 125.0 m away, it has risen vertically by a height of 5.7 m and slowed down to a speed of 6.5 m/s.What is the force of friction on the car?

Solution:

Etotal before = Etotal after

Ek + Ep = Ek' + Ep' + Eheat

0.5 mv2 + mgh = 0.5 mv'2 + mgh' + Eheat

Eheat = 0.5 mv2 + mgh - 0.5 mv'2 - mgh'

Eheat = 0.5 x 1700 x 13.92 + 1700 x 9.81 x 15.7 - 0.5 x 1700 x 6.52 + 1700 x 9.81 x 10.0

Eheat =2.2 x 105 J

F x d = Eheat

F = 2.2 x 105 Ö 125

F = 1.76 x 103 N

 

6.   What is the maximum speed that the 10 kg ball on the pendulum in the diagram shown at right will achieve if it is released from a height of 5.0 m above its lowest point?

Solution:

Ek = Ep

0.5 mv2 = mgh           v = 10 m/s

 

7.   A pole vaulter of mass 75 kg just clears 5.3 m.  How fast was he running the instant before his jump, assuming the height he jumps is due entirely to his kinetic energy?

Solution:

Ek = Ep

v = Ã(2gh)

v = 10.2 m/s

 

8.   How much energy has been lost due to frictional heating by the air when an 0.08 kg object attains a speed of 5.0 m/s while free falling 1.5 m from rest?

Solution:

Use the conservation of energy law

Total Energy before = Total Energy after

P.E. = K.E. + heat energy
mgH = 0.5mv2 + heat energy
heat energy = 0.20 J

 

9.   Explain what happens to the PE, KE  and total Energy of a brick falling from a chimney.

Solution:

At the start the brick has Potential Energy and no Kinetic Energy, as the brick falls its Potential Energy is being converted into KE (mgÆh = 0.5 mv2).  The total energy at any point in time during the fall is constant (mgÆh + 0.5 mv2 = mgH).  At the bottom of the fall, all the Potential energy has been converted into Kinetic energy (0.5 mv2 = mgH)

 

10. A biker approaches a hill with a speed of 8.5 m/s.  The total mass of the bike and rider is 85 kg. 

a) Find the kinetic energy of the bike and rider.

b) The rider coasts up the hill.  Assuming there is no friction, at what height will the bike come to a stop?

c) Does your answer depend on the mass of the bike and rider?  Explain.

Solution:

a) Ek = 0.5 mv2         b) Ep = Ek     c) No

    Ek = 0.5 x 85 x 8.52               mgh = 3070                 mgh = 0.5 mv2

    Ek = 3070 J                h = 3070 / (85 x 9.8)              gh = 0.5 v2

          h = 3.7 m                 h = 0.5 v2 / g

 

11. Betty weighs 420 N and is sitting on a playground swing seat that hangs 0.40 m above the ground.  Tom pulls the swing back and releases it when the seat is 1.00 m above the ground.

a) How fast is Betty moving when the swing passes through its lowest position?

b) If Betty moves through the lowest point at 2.0 m/s, how much work was done on the swing by friction?

Solution:

mg = 420, m = 42.8 kg

 

a) Ep = Ek     b) W = ÆE

    mgh = 0.5 mv2   where h = height fallen                  W = Ekf - Epi

    2gh = v2          W = 0.5 x 42.8 x 22 - 420 x 0.6

    Ã(2gh) = v        W = 85.6 - 252

    Ã(2 x 9.81 x 0.6) = v                 W = 166 J lost to heat energy

    v = 3.43 m/s