Answers
These are the topics for the questions in this unit.
Work and Power
(back
to the questions)
(back to the top)
1. Work is the product of a force and the displacement
over which that force is applied.
The force and displacement
must be along the same line.
2. Yes. Both by the ground pushing the car
forward and the resistive forces pushing
on the car against its motion.
3. No. The average force applied to an object
to lift it equals its weight no matter how
fast the lift is done, so the work is the same.
4. Since friction opposes the motion of the object,
it is in the opposite direction of the
displacement. Therefore
it would be a negative force, doing negative work.
5. a) W = F d = 50 N (10 m) = 500 Nm
b) No. The normal
force is perpendicular to the displacement.
c) At constant speed the
box is in equilibrium. Ff = -50 N W = -50
N (10 m)
= -500 Nm
6. In this case the 50 N is not along the same line
as the displacement. You have to
use the component of the 50
N force which is along the displacement.
Fh = 50 N (cos 30°) = 43 N
W = 43 N (10 m) = 430 Nm
7. Power is the rate at which work is done. P = W / T
8. P = W / T = 2000 J / 30 s = 67 W
Potential Energy (back to the questions) (back to the top)
1. Potential energy is the ability to do work an object possesses due to its position.
2. a book on the shelf (gravitational potential
energy)
a loaded dart gun (elastic
potential energy)
a stick of dynamite (chemical
potential energy)
a chunk of Uranium (nuclear
potential energy)
a battery (electrical
potential energy)
3. Since gravitational potential energy is calculated
by the product, m g h, doubling
the height doubles the gravitational
potential energy.
4. a) PE = m g h = 30 N (0 m) = 0 J
(30 N is the weight of the book = m g)
b) PE = m g h = 30 N (1.5
m) = 45 J
c) PE = m g h = 30 N (-
2.0 m) = - 60 J
5. PE = m g h h = PE / mg
= 7.5 J / (0.25 Kg (9.8 m/s2 )) =
3.0 m
Kinetic Energy (back to the questions) (back to the top)
1. Kinetic energy is the ability of an object to do work due to its state of motion.
2. Since kinetic energy is calculated by the product,
(1/2 ) m v2 , doubling the speed
will increase the KE by a factor
of 4. (22 )
3. No. Since yours has twice the speed it
will have four times the energy. (See #2)
Since height is directly related to energy, your ball will go four times
as high.
4. KE = (1/2 ) m v2 = (1/2 ) (10 Kg) (4 m/s)2 = 80 J
5. KE = (1/2 ) m v2 v = sqrt(2 KE / m) = sqrt(2 (100 J / 0.045 Kg) = 67 J
6. KEcar = KEbullet
(1/2 ) mcVc2
= (1/2 ) mbVb2
(1/2 ) (1000 Kg) Vc2 = (1/2
) (0.050 Kg) (400 m/s)2
Vc = 8.0 m/s
Change in Energy (back to the questions) (back to the top)
1. The connection between force and energy is work.
W = D E. In this problem,
W =
D
KE F
d = KEf - KEi
where KEi = (1/2 ) (0.030 Kg) (400 m/s)2
F (0.04 m) = 0 J - 2400 J
= 2400 J
F = 60,000 N
KEf = 0 (it stopped)
The energy of the bullet is converted to internal energy of the bullet
and the tree
(their temperature increases, they are deformed) and also converted
into sound.
2. No. Since the barrel of the rifle is longer,
the expanding gases will do more work
on the rifle bullet and it will have more KE. This means it will
move faster.
3. If you assume the force of the brakes is the
same in both cases, the stopping distance is
directly related to the amount
of KE to be lost. Since KE depends on V2
the ratio of the
stopping distances is the same
as the ratio of the V2 s.
(25mph)2 / (15 mph)2
= 2.8 It will take 2.8 times as far to stop.
4. a) Yes. It will have less gravitational
potential energy.
b) The energy goes into
what ever stopped the elevator, increasing its temperature,
as well as creating sound.
5. a) Internal. Without the gravitational
force, the object would not have any potential
energy in the first place.
b) No. It is converted
from gravitational potential into kinetic energy. The total
energy only changes after it has landed.
6. a) Yes. Since velocity is a vector
quantity the change of direction is the important
factor. D
V = Vf - Vi = - 170 mph
b) the bat
c) The work done is a
product of F d. A stronger batter could apply a larger F and a
proper follow through would increase d. In both cases more work would
be
done causing a greater * KE.
7. W = D E in this case W = D KE F d = D KE = 500 N (3 m) = 1500 J
8. D PE = PEf
- PEi = mghf - mghi
= mg (hf -hi ) = 0.2 Kg (9.8 m/s2
) (0 - 1.2 m)
= - 2.4 J
9. W = D PE =
PEf - PEi = mghf - mghi
= mg (hf -hi ) = 15 Kg (9.8 m/s2
) (2 m - 0)
= 294 J
10. The work done on the dart gun will become the
kinetic energy of the dart if we assume no
energy is lost in
the firing process. (in reality some energy is lost to friction and
sound)
W = D KE
F d = KEf - KEi
16 N (0.08 m) = (1/2 ) 0.020 Kg (V)2
= KEf
11.3 m/s = V
11. Work done by the motor becomes the energy of
the train. Since we are ignoring the
small KE,
W = D PE.
W = D PE
since P = W / T , W = P *T So,
P DT = D
PE
P DT = PEf - PEi
= mghf - mghi
= mg (hf -hi )
44,000 W (20 s) = 3000 Kg (9.8 m/s2
) (hf - 0)
30 m = hf
12. Since P = W / T and
W = D KE P =
D
KE / T = (KEf - KEi ) / T
= [(1/2 ) (0.3 Kg) (30 m/s)2
- 0] / 0.75 s
= 300 J
Conservation of Energy (back to the questions) (back to the top)
1. No. The velocity of the car definitely
changed. The energy was transformed into
internal energies of the car and the tree (their temperatures increased,
their
shapes were deformed) and sound was created.
2. KEmin at the ends of the swing,
KEmax at the bottom of the swing.
PEmin at the
bottom of the swing, PEmax at the ends of the swing.
3. When you are pulled out at an angle on the swing
you are higher than at the bottom.
As you swing, your loss of PE
becomes a gain in KE.
D
PE = D KE
PEbottom - PEtop
= - (KEbottom - KEtop)
(opposite sign because one is a loss and the other is a gain)
PEtop = KEbottom
(since PEbottom= 0 and
KEtop= 0)
mghtop = (1/2 ) mV2bottom
Vbottom = sqrt(2ghtop)
Since the mass cancels out, you and your cousin will go the same speed
at
the bottom.
4. Using the same reasoning as in #3,
Vbottom = sqrt(2ghtop)
= sqrt(2(9.8 m/s2 (0.5 m) )
= 3.1 m/s
5. For similar reasoning as in #’s 3 and 4,
KEbottom
= PEtop
(1/2 ) m V2 = m g htop
(1/2 ) V2 = g htop
htop = (1/2 ) V2
/ g
htop = (1/2 ) (30 m/s)2/
9.8 m/s2
= 46 m
6. Use the same logic as in #5. This time solve for V.
(1/2) V2 = g htop V = sqrt(2g htop) = sqrt(2(9.8 m/s2 ) (12 m)) = 15 m/s
7. a) Vbottom
= sqrt(2ghtop ) = sqrt(2(9.8 m/s2
) (35 m)) = 26 m/s
b) No. The mass
does not appear in the speed equation.
c) Assuming no energy
is lost between the top of the first hill and the second hill,
the total energy would not change.
TEtop = TEhill
PEtop + KEtop
= PEhill + KEhill
mghtop + 0 = mghhill
+ (1/2) mV2hill
ghtop + 0 = ghhill
+ (1/2) V2hill
(since all terms have
9.8 m/s2 (35 m) = 9.8
m/s2 (15 m) + (1/2) V2hill
mass it cancels out)
20 m/s = V
8. Since some energy will be lost to friction along
the ride, the lower height of the corkscrew
assures the train will go through
the loops with enough speed so you won’t fall out of the seat.
9. Reference #7 for the logic of the following equation.
ghtop + 0 = ghloop
+ (1/2) V2loop
(with no loss of energy)
9.8 m/s2 (20 m) = 9.8
m/s2 (12 m) + (1/2) V2loop
12.5 m/s = V
TE = PEloop + KEloop
= 9.8 m/s2 (12 m) + (1/2) (12.5 m/s)2
= 196 J (theoretical)
= 9.8 m/s2 (12 m) + (1/2) (10.0 m/s)2
= 168 J (actual)
% energy lost = (theoretical - actual) / theoretical = (196 J - 168 J) / 196 J = 0.14
The train loses
14 % of its energy before getting to the top of the loop.
10. a) At the top of the trajectory after
each bounce TE = PE + KE , but KE = 0 .
So, TE = PE = mgh.
The % energy lost = (Ebefore - Eafter
) / Ebefore
= (mghbefore - mghafter
) / mghbefore = (hb
- ha ) / hb
For the first bounce, (2.5 m - 2.1 m) / 2.5 m = 0.16 = 16 %
b) The energy
was transformed into internal energy of the floor and ball and
created sound.
c) If the ball
loses 16 % of its energy on each bounce, it will come up to 84 %
(0.84) of the height it had before the bounce.
After the first bounce it comes back to..........0.84 (2.5 m) = 2.1 m
After the second bounce it comes back to.......0.84 (2.1 m) = 1.76 m
After the third bounce it comes back up to.....0.84 (1.76 m) = 1.48 m