Download Practice Problems with Solutions

Practice Problems with Solutions Based Chapter 8 and 9, and good review for Midterm 1.
Physics 11a Fall 2010
1. You push on a crate, and it starts to move but you don’t. Draw a free-body diagram for
you and one for the crate. Then use the diagrams and Newton's third law of motion to
explain why the crate moves but you don't.
1. Sol’n:
Newton’s third law tells us that the force I exert on the crate and the force the crate exerts
on me have equal magnitudes and opposite directions. That the crate moves and I don’t
implies that the sum of all the forces acting on the crate is non-zero while the sum of all
the forces acting on me is zero. That is, the floor must be exerting a greater frictional
force on me than on the crate.
2. In the figure below, a 50-kg skier heads down a slope, reaching a speed of 35 km/h.
She then slides across a horizontal snow field but hits a rough area. Assume the snow
before the rough area is so slippery that you can ignore any friction between skis and
snow. If the frictional force exerted by the snow in the rough area is 40 N, how far across
the rough area does she travel before stopping?
Figure for Problem 2
2. Sol’n: A constant force implies a constant acceleration, for which we know the
equations of motion.
F !40 N
a= =
= !0.80 m/s 2
m 50 kg
v(t) = vi + at = (35 km/h)
1 h 1000 m
+ (!0.80 m/s 2 )t = 0 m/s
3600 s 1 km
t = 12 s
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3. The gravitational force acting on the piano in the figure is 3000 N. With how much
force do you have to pull on the rope in order to hold up the piano (a) with the single
pulley shown on the left and (b) with the block-and-tackle system shown on the right?
Figure for Problem 3
3. Sol’n: (a) The forces acting on the piano are just the force of gravity and the tension
in the rope. Since the piano is not accelerating, the tension in the rope must have the
same magnitude as the force of gravity acting on the piano. I have to pull against that
same tension, namely, with 3000 N of force.
(b) Here, the forces acting on the piano are the force of gravity and twice the tension in
the rope. That is, the rope pulls on both sides of the lower pulley with its tension. So, I
only have to pull with one half of the piano’s weight, namely, 1500 N.
4.(II) Two blocks are tied together by a string draped over a pulley, as shown in Figure
8.15. (a) At first, the blocks don’t slip. How does the tension in string compare with the
gravitational force exerted on the dangling block? (b) You tip the table a bit, and the
blocks begin to slip. Now how does the tension in the string compare with the
gravitational force exerted on the dangling block?
Figure for Problem 4
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4. Sol’n: (a) If the blocks don’t slip, then the their acceleration is zero, so the net force
on the dangling block must be zero, too. The only forces acting on the dangling block are
the tension in the string and the force of gravity, so the tension in the string must be equal
in magnitude to the gravitational force on the dangling block.
(b) Since the dangling block is accelerating downward, the net force acting on it must
also be directed downward. So, the tension in the string must be less than the
gravitational force on the dangling block.
5. All the blocks in the figure have the same inertia. Rank the different configurations in
increasing order of tension in the rope. (Hint: free-body diagrams are a good place to
start.)
Figure for Problem 5
5. Sol’n: Because nothing is moving in (a), (b), and (c), which means that the
acceleration is zero, it’s easy to see that the tension in the rope exactly balances the force
of gravity acting on each object. If the inertia of each block is m, then the tension equals
mg, mg, and 2mg, in (a), (b), and (c), respectively.
In (d), the forces acting on the objects on each side are the force of gravity and the
tension in the rope, and the rope ensures that both objects accelerate at the same rate.
Combining these, we get:
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F T ! mg = ma
(2m)g ! F T = (2m)a
2F T ! 2mg = 2mg ! F T
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F T = mg
3
So, the order is (a) = (b) < (d) < (c).
6. Two springs have spring constants k1 and k2, with k1 < k2. Connected as shown in the
figure, they act like one spring. What is the spring constant of the combination: (a)
smaller than k1, (b) equal to k1, (c) between k1 and k2, (d) equal to k2, (e) greater than k2?
Explain your answer.
Figure for Problem 6
6. Sol’n: In order to stretch the pair of springs a given distance, we have to stretch each
spring by that distance. That is, we have to provide a force equal to the sum of the forces
exerted by each spring.
F = k!x = k1!x + k2 !x = (k1 + k2 )!x
k = k1 + k2 > k2
That is, the answer is (e).
7. A modified Atwood machine is shown in the figure below. Each of the three blocks
has the same inertia m. One end of the vertical spring, which has a spring constant k, is
attached to the single block, and the other end of the spring is fixed to the floor. The
positions of the blocks are adjusted until the spring is at its equilibrium length. The
blocks are then released from rest. What is the acceleration of the two blocks on the right
after they have fallen a distance D?
Figure for Problem 7
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7. Sol’n: The forces acting on the block on the left are the force of gravity, the force
exerted by the spring, and the tension in the rope. The forces acting on the pair of blocks
on the right are the force of gravity, and the tension in the rope. The rope ensures that
both sides accelerate at the same rate. Combining these, we get:
F T ! kD ! mg = ma
(2m)g ! F T = (2m)a
2F T ! 2kD ! 2mg = 2mg ! F T
2kD + 4mg
FT =
3
T
F
g kD
a=g!
= !
2m 3 3m
8. A 1.2-kg ball dropped from a height of 3.0 m onto a steel plate rigidly attached to the
ground bounces back up to a height of 2.5 m. (a) What is the impulse delivered to the ball
by the steel plate? (b) What is the coefficient of restitution of the collision?
8. Sol’n: (a) The impulse delivered to the ball equals the change in its momentum. We
can calculate its momentum from its kinetic energy, which, because of conservation of
energy, equals the gravitational potential energy it has at the top of its trajectory.
p2
K=
= mgh
2m
p = 2m2 gh
J = !p = p f + pi
J = 2(1.2 kg)2 (9.8 m/s 2 )(2.5 m) + 2(1.2 kg)2 (9.8 m/s 2 )(3.0 m) = 18 kg " m/s
(b) The coefficient of restitution equals the ratio of the final velocity to the initial
velocity. This is also equal to the ratio of the final momentum to the initial momentum,
and the square root of the ratio of the final height to the initial height.
e=
vf
vi
=
mv f
mvi
=
pf
pi
=
2m2 gh f
2m2 ghi
=
hf
hi
=
2.5 m
= 0.91
3.0 m
9. Does work done on a system necessarily change the system's kinetic energy?
9. Sol’n: No. Any work done on the system can change other forms of energy in the
system. The work done in squeezing a soft rubber ball goes into changing the
deformation energy of the system, for instance.
10. A block is launched at speed v, slides down an incline (there is friction between the
block and incline), and into a spring (Figure 9.25) where it slows to zero speed as the
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spring compresses. (a) Draw an energy diagram for this process using a system consisting
of the block, spring, Earth, and incline. If there is more than one object causing energy
transformations, use a separate bar for each. (b) Repeat using a system of just the block
and incline. (c) Why can’t you use the block alone as your system?
10. Sol’n (a) For the system of block, spring, Earth and incline, all forces are internal
(isolated system), but energy is dissipated by friction (not a closed system). Thus no work
is done. The energy diagram involves changes in kinetic energy, gravitational potential
energy, elastic (spring) potential energy, and thermal energy as shown in figure (a)
below. Note that we do not have enough information to decide the relative sizes of each
energy change; we know only that the sum of losses in kinetic and gravitational potential
energies must equal the gains in elastic potential energy and thermal energy.
(b) For the system of block and incline, the external gravitational force exerted by the
Earth and the external elastic force exerted by the spring each do work on the system.
There is a change in kinetic energy and thermal energy, as shown in figure (b).
ΔK
ΔUs
ΔUg
ΔEs
ΔEth
Ws
Wg
ΔEth
Ws
Wg
(a) System of block, spring, Earth and incline
ΔK
ΔUs
ΔUg
ΔEs
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(b) System of block and incline
(c) If we choose the block alone as our system, we don’t know how much of the
energy dissipated by friction winds up as thermal energy in the block; we only know the
total amount that winds up as thermal energy distributed between the block and the
inclined plane.
11. You throw a ball straight up into the air. If you consider air resistance, which takes
more time: the upward trip or the downward trip? Analyze using three different systems.
11. Sol’n: Whenever the ball is moving, air resistance acts to decrease its kinetic energy.
(a) Using the ball as our system, there are two external forces doing work: gravity and air
resistance. While we don’t know how much of the resulting thermal energy winds up in
the system (ball) and how much of it winds up in the air, we do know that the ball’s
kinetic energy decreases. So, if the ball starts with some initial kinetic energy, travels to
the top of its path, and falls back to its initial height (which means that gravity will have
done equal amounts of positive and negative work during the round trip, yielding zero
work on the ball), it will have less kinetic energy than it had when it started. That is, its
speed at the end of the trip will be less than its initial speed was. Since this is true for any
such segment of the ball’s trip through the air, its speed at any given height above the
ground is less on the way down than it was on the way up. Since the distance covered is
the same in both cases, the trip down must take longer. (b) Using the ball+air as our
system, the thermal energy dissipated by air resistance remains inside the system. The
external force of gravity does zero work for the round trip, so any increase in thermal
energy must come at the expense of kinetic energy. The result is the same: the trip down
takes longer. (c) Using the ball + air + Earth, all forces are internal and no work is done.
The gravitational potential energy changes by zero for the round trip, but as before there
is an increase in thermal energy of the system. This comes at the expense of kinetic
energy, yielding the same result.
12. At the end of a delivery ramp, a skid pad exerts a constant force on a package so that
the package comes to rest in a distance d. The ramp is changed so that the same package
arrives at the skid pad at a higher speed and the stopping distance is 2d. What happens to
the length of time it takes for the package to stop?
12. Sol’n: The amount of energy dissipated by a constant force equals the magnitude of
the force times the distance traveled, so the package must be arriving with twice the
kinetic energy in order for the stopping distance to be doubled, which means that its
velocity must have been increased by a factor of the square root of 2.
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!"K = "Eth = Fd
K 2 = 12 mv22 = F(2d) = 2(Fd) = 2K1 = 2 12 mv12
v2 = 2v1
The same force in both cases implies the same acceleration, so the amount of time
required for the package to come to a stop must increase by the same factor.
!v
a=
!t
!t2 =
!v2
=
a
2!v1
!v
= 2
= 2!t1
a
a
13. A beetle that has an inertia of 4.0 × 10–6 kg sits on the floor. It jumps by using its
muscles to push against the floor and raise its center of mass. (a) If its center of mass
raises 0.75 mm while it is pushing against the floor and then continues to travel up to a
height of 300 mm above the floor, what is the magnitude of the force exerted by the floor
on the beetle? (b) What is the beetle’s acceleration during the launch?
13. Sol’n: (a) When the beetle rises to a height of 300 mm, the kinetic energy it has as it
leaves the floor is converted into gravitational potential energy. That is, from the height
to which it rises, we know how much kinetic energy it has as it leaves the ground. That
kinetic energy is equal to the product of the net force acting on it, as it was pushing off,
times the displacement of its center of mass while the force was acting on it. The net
force equals the force of gravity, which turns out to be negligible, plus the contact force
exerted by the floor.
K = mg!h = (4.0 " 10#6 kg)(9.8 m/s 2 )(0.300 m # 0.00075 m) = 1.2 " 10#5 J
K = Fext !xcm
Fext =
K
1.2 " 10#5 J
=
= 0.016 N
!xcm 0.00075 m
Fext = FEbG + Fgbc = #mg + Fgbc
Fgbc = Fext + mg = 0.016 N + (4.0 " 10#6 kg)(9.8 m/s 2 ) = 0.016 N
(b) The beetle’s acceleration equals the force acting on it divided by its inertia.
F
0.016 N
a= =
= 3.9 ! 103 m/s 2
-6
m 4.0 ! 10 kg
Note that we retained a few extra digits in the intermediate results to arrive at this answer.
14. At time t0, you are sitting on a dock and you begin using a rope to lower a 4.0-kg
lobster trap until at time t1 it just reaches the water surface 1.4 m below the dock. You
then let the trap sink to the bottom of the ocean, which it reaches at time t2. The trap is
falling at a constant 1 m/s when you starts lowering it from the dock at t0. Let your
system be (Earth + trap). (a) Do you do any work on the trap in the interval from t0 to t1?
(b) What forces act on the trap during the interval from t0 to t1? (c) Draw the work-energy
diagram for the interval from t0 to t1, showing a correct numerical scale.
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14. Sol’n:
(a) Yes, because, even though the Earth-trap system is losing gravitational
potential energy, it’s not gaining kinetic energy. That is, you must be doing negative
work.
(b) The tension in the rope, due to the your hands, is the only external force
acting on the system. The force of gravity acts on the trap, too, but that force is internal
to the system and, so, does no work on the system.
(c) The Earth-trap system loses (4.0 kg)(9.8 m/s2)(1.4 m) = 55 J gravitational
potential energy as the trap is lowered to the water, and you do exactly that much
(negative) work.
15. In the block-and-tackle arrangement shown in the figure, the single rope pulls on the
block three times. (a) Show that the magnitude Fcrb of the pulling force needed to raise
the block at constant speed is ⅓(mg), where m is the inertia of the block. (b) One worker
uses this block-and-tackle arrangement to lift a heavy block, while another hauls up an
identical block with a straight rope. After both blocks have been raised to the same
second-floor window, has either worker done more work than the other? If so, which
worker has done more work?
Figure for Problem 15
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15. Sol’n: (a) Since the block is rising at a constant speed, the vector sum of the forces
acting on it must be zero. Those forces are three times the tension in the rope, pulling it
up, and the force of gravity, pulling it down.
3T = mg
T = 13 mg
Alternatively, in terms of work, the block rises by one-third the distance the rope is
pulled.
Fpulling d pulled = mg!h = mg( 13 d pulled )
Fpulling = 13 mg
(b) Ignoring friction, both do the same amount of work, because both blocks have been
raised through the same distance against the same force of gravity. Alternatively, we get
the same answer by noticing that the worker who uses the block and tackle pulls his rope
three times the distance, against one-third the force, as the other worker does. We might
expect that there’d be more friction with the block and tackle, though, in which case a
little more work would be done in using it.
16. A cart of inertia mcart sitting on an air track is attached to a string that leads over a
pulley to a hanging block of inertia mblock. Use the energy law to determine the speed of
the cart after the block is freed and allowed to fall and the cart has gone a distance d
along the track.
16. Sol’n: The change in energy of the system equals the amount of work performed on
it. The change in the cart-block system’s energy is just the kinetic energy it gains, and
the work done on it is just the work done by the force of gravity as the block falls.
Putting these together, we have
!E = W
W = Fext d = mblock gd
!E = K = 12 (mcart + mblock )v 2
1
2
(mcart + mblock )v 2 = mblock gd
v=
2mblock gd
mcart + mblock
17. A 60-kg ice skater stands facing a wall with his arms bent and then pushes away from
the wall by straightening his arms. His center of mass has moved 50 cm at the instant at
which his fingers lose contact with the wall, and at this instant he is traveling at 3 m/s.
What are (a) the average force exerted by the wall on him, the work done by the wall on
him, and (c) the change in the kinetic energy of his center of mass? [RB10-56]
17. Sol’n: (a) The change in the kinetic energy of his center of mass equals the product
of the external force acting on him and the displacement of his center of mass.
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Fext !xcm = !K cm
!K cm 12 (60 kg)(3.0 m/s)2
270 J
=
=
= 5.4 " 102 N
!xcm
0.50 m
0.50 m
(b) The work done by the wall equals the product of the force exerted by the wall times
the displacement of the point of application of the force. Since the wall doesn’t move,
the displacement of the point of application of the force equals zero, so the work done by
the wall equals zero, too.
(c) As we calculated in (a), the change in kinetic energy of his center of mass equals
2.7 ! 102 J .
Fext =
18. A 6.0-kg baby is held just above a mattress and then released from rest and allowed to
fall, sinking into the mattress. Model the mattress as a single spring with no inertia and k
= 500 N/m. (a) Find baby’s kinetic energy, the gravitational potential energy of the
system (baby, spring and Earth), and the elastic potential energy at the following
compressions: 0.0 m, 0.050 m, 0.10 m, 0.15 m, 0.20 m. (b) What is the maximum
compression of the mattress?
18. Sol’n: If we choose our system to be the baby, the Earth, and the spring, we have a
closed system, so conservation of energy applies. As the baby falls, its kinetic energy
increases, the gravitational potential energy decreases, and the potential energy stored in
the spring increases.
When the compression of the mattress is 0.0 m, that is, just before we release the
baby, the baby is at rest, so its kinetic energy is zero, and the spring is not compressed at
all, so the potential energy stored in it is zero. We’re free to choose any value for the
gravitational potential energy, here, since it’s only change in potential energy that
matters, so let’s choose it to be zero, too.
!E = !K + !U G + !U s = 0
K(x) = U G (x) " U s (x)
K(x) = mgx " 12 kx 2
x = !0.050 m
U G = (6.0 kg)(9.8 m/s 2 )( ! 0.050 m) = !2.9 J
U s = 12 (500 N/m)(!0.050 m)2 = 0.63 J
K = 2.3 J
x = !0.10 m
U G = (6.0 kg)(9.8 m/s 2 )( ! 0.10 m) = !5.9 J
U s = 12 (500 N/m)(!0.10 m)2 = 2.5 J
K = 3.4 J
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x = !0.15 m
U G = (6.0 kg)(9.8 m/s 2 )( ! 0.15 m) = !8.8 J
U s = 12 (500 N/m)(!0.15 m)2 = 5.6 J
K = 3.2 J
x = !0.20 m
U G = (6.0 kg)(9.8 m/s 2 )( ! 0.20 m) = !12 J
U s = 12 (500 N/m)(!0.20 m)2 = 10 J
K = 1.8 J
(b)
When the mattress is maximally compressed, the baby’s speed will be
(instantaneously) zero, so its kinetic energy will be zero, too. That is, the gravitational
potential energy lost since the start of its descent will equal the potential energy stored in
the spring.
!mgx = 12 kx 2
x=!
2mg
2(6.0 kg)(9.8 m/s 2 )
=!
= !0.24 m
k
500 N/m
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