Download Physics 141H Homework Set #3 Chapter 3: Multiple

Physics 141H
Homework Set #3
Chapter 3:
Multiple-choice:
1) Newton’s First Law tells us that if no force acts on an object, its velocity will never
change. (D)
3) With the band stretched to the 7in mark , the block has not yet started to move. This
means its acceleration is 0, so by Newton’s Second Law the net force on the block is 0.
The same was true when the band was stretched only to the 6in. mark. (B)
7) Newton’s Second Law says that the acceleration of an object must be in the same
direction as the net force acting on it. (B)
9) Newton’s Third Law tells us that forces always occur in pairs, with equal magnitudes
(but opposite directions!). (B)
16)
a) In the case where you’re standing in the bathroom, there are two forces acting on you –
your weight, pulling you down, and the scale pushing up on your feet. The spring in the
scale compresses by an amount proportional to this second force, and that corresponds to
the reading on the dial. So, the best answer is perhaps (D). But, since this force is equal
to your weight, and also (by Newton’s Third Law) equal to the force of your feet pushing
down on the scale, (B) and (C) are also true.
b) Now that you’re in an elevator accelerating upwards, the net force on you must be
upwards as well. This means that the force of the scale pushing up on your feet is greater
than your weight. (C) and (D) are both true in this case.
Questions:
10) There is no relation between the two. The net force on an object is always in the
same direction as the acceleration, but that tells us nothing about the direction of the
velocity.
12) I would point out that there is more than one force acting on the horse, as shown
below:
Fc
Fg
Fg
The cart pulls backward on the horse with force FC. But, if the horse pushes its hooves
backwards against the ground, the ground will, by Newton’s Third Law, push forward on
the horse. I’ve indicated these forces with Fg. If the Fg’s add up to a force greater than
Fc, the horse can indeed move forward.
19) Yes. If the body is resting on a surface with very little friction, a small horizontal
force can cause it to start moving.
27) They may measure different velocities if the frames are moving with respect to one
another. Since the acceleration is the same in all inertial frames, and the mass is a
property of the object itself and not the reference frame, both observers will measure the
same force acting on the object.
34) The spring scale measures the force between itself and the hanging weight. If (a) the
elevator is at rest, the force of the scale pulling up on the weight exactly balances the
force of gravity pulling it down. This is also true for (b), the elevator moving with
constant velocity. In case (c) the elevator descends with decreasing speed, so its
acceleration is upwards. For the weight to also accelerate upwards, the scale must supply
a force greater than gravity. Finally, in case (d) the acceleration is downwards, so the
scale provides a force less than that of gravity.
Therefore, the scale has the largest reading in case (c), and the smallest in case (d).
Exercises:
5) We start by drawing a force diagram for the sled:
F1 = 92N
Sled
F2 = 90N
Ice
To find the net force on the sled, we take the x axis as horizontal, and write the force
vectors in terms of their components:
F1 = −92Ni + 0Nj
F2 = 90Ni + 0Nj
Adding these, we find the net force to be −2Ni , which has magnitude of 2N. Thus the
magnitude of the acceleration is:
a =
F
2N
m
=
= 0.08 2
m 25kg
s
Note that we weren’t asked to find the direction of the acceleration, and indeed weren’t
given enough information to do so!
10)
a) We want to go from a velocity of 0 to 1/10th the speed of light, or 3 x 107 m/s, in a
given amount of time. Using the equation ∆v = at , we can find the acceleration required:
a=
∆v v f − vi v f
=
=
t
t
t
Newton’s Second Law then tells us that the force required is:
F = ma =
mv f
t
Plugging in the numbers, we get:
1. t = 3 days:
(1200 metric tons) 3 × 107
F=
m
s
3 days
1000kg
1 day
1 metric ton 1.7 × 105 s
= 7.1 × 10 N
7
2. t = 2 months:
F=
(1200 metric tons) 3 × 107
2 months
= 3.5 × 10 N
6
m
s
1000kg
1 month
1 metric ton 5.1 × 106 s
b) The distance covered during the acceleration phase is given by x =
vf
1 2
at . But we
2
1
v f t . In other words, the total distance traveled during
t
2
the acceleration is ½ of that traveled in the same amount of time after the final velocity is
reached.
know that a =
, so that x =
In this case, a final velocity of 1/10th the speed of light means that it takes 10 months to
travel one light-month. So, in the case where 3 days are spent accelerating, 0.15 lightdays, or 0.005 light months, are covered during acceleration. Given the number of
significant figures we have to work with, there are still 5 light-months of distance to go,
so the trip takes 50 months.
In the case where two months are spent accelerating, 0.1 light-months are covered during
acceleration, leaving 4.9 light-months to go. So, the total time is 49 months at the final
velocity, plus the 2-month acceleration time, for a total of 51 months.
17) The automobile has a weight of 3900 lb. This means that gravity exerts a force of
3900 lb on the car (remember that pounds are a unit of force, not mass!). We also know
that any object falling due to gravity has an acceleration of g = 9.8m/s2 = 32 ft/s2. In this
problem the car is accelerating at 13 ft/s2, or 0.41g.
By Newton’s Second Law, this means that the net force on it must be 0.41times its
weight, or 1600 lb.
19) The net force on the plane comes from the two engines. Since each engine provides a
thrust of 1.40 x 105 N, the total force is 2.80 x 105 N. We can then find the mass of the
plane:
m=
F 2.80 × 105 N
=
= 1.22 × 105 kg
2
a
2.30 m/s
But the problem asked for the weight of the plane, not the mass, so the answer is:
W = mg = (1.22 × 105 kg )( 9.8m/s 2 ) = 1.20 × 106 N
23)
a) The first thing we need to do is find the man’s velocity at the moment his feet touch
the ground. We know that he jumped from a height of 0.48m, so we can find the time it
took him to fall:
1 2 1 2
at = gt
2
2
2 ∆y
t=
g
∆y =
From this we can find the velocity with which he hits the ground:
v = at = gt = g
2 ∆y
= 2 g ∆y
g
= 2 ( 9.8m/s 2 ) ( 0.48m ) = 3.07m/s
Now consider his motion after his feet hit the ground. He decelerates from 3.07m/s to
0m/s while traveling a distance of 2.2 cm. So we have:
∆v = at
t=
∆v
a
1
1 ∆v
∆y = at 2 = a
2
2
a
2
=
( −3.07m/s )
1 ∆v 2
1
=
a=
2 ∆y
2
0.022m
1 ∆v 2
2 a
2
= 214m/s 2
b) The force acting on the man as he decelerates is given by Newton’s Second Law:
F = ma = (83kg ) ( 214m/s 2 ) = 1.77 × 104 N = 3.9 × 103 lb
Ouch!
28) To determine the force we need to apply to the plane, we first must find the required
acceleration. We know that the plane starts at rest, and must attain a velocity of 280ft/s
while traveling 300ft along the flight deck. So we have:
∆v = at
t=
∆x =
∆v
a
1 2 1 ∆v
at = a
a
2
2
1 ∆v 2
1
a=
=
2 ∆x
2
2
=
( 280ft/s )
300ft
1 ∆v 2
2 a
2
= 131ft/s2
The plane weighs 26 tons or 5.2 x 104 lb, which means its mass is:
m=
W 5.2 × 104 lb
=
= 1.6 × 103 slug
2
g
32ft/s
Thus the total force we need to apply to the plane is
F = ma = (1.6 × 103 slug )(131ft/s2 ) = 2.1 × 105 lb
The plane’s engine supplies 2.4 x 104lb, meaning that the catapult must add 1.9 x 105 lb.
Problems:
3) We have the following situation:
y
16 m
x
The first block starts from rest, moves down the ramp with constant acceleration, and
reaches the bottom, 16m away, in 4.2s. It is most convenient to define the coordinate
system as shown above, so that we can consider motion only in x.
a) Using x =
1 2
at for the first block, we find that
2
a=
2 x 2 (16m )
=
= 1.8m/s 2
2
t2
( 4.2s )
Since we are told that both blocks experience the same acceleration, this is also the
acceleration of the second block.
b) The second block starts at the bottom of the ramp with some initial velocity vo, goes
some distance up the ramp, then slides back to the bottom. The time this round-trip takes
is exactly as long as it took the first block to slide to the bottom, or 4.2s. We also know
(from part a) that the block accelerates at 1.8m/s2 .
From the symmetry between the block’s trip up the incline and the trip back down, we
know that the final velocity when it returns to the bottom of the ramp has the same
magnitude as the initial velocity, but in the opposite direction. So,
v f − vo = − vo − vo = at = (1.81m/s2 ) ( 4.2s ) = 7.6m/s
-2vo = 7.6m/s
vo = −3.8m/s
The minus sign makes sense, since we took the positive x direction to be down the ramp.
c) The second block reaches its highest point halfway through the round-trip (again,
symmetry tells us this). So we can find how far it goes by using:
1 2
at + vo t
2
1
2
= (1.81m/s 2 ) ( 2.1s ) + ( −3.8m/s )( 2.1s )
2
= −4.0m
∆x =
Again, the minus sign makes sense because the block moves closer to the origin on its
upward trip. The answer we’re looking for is that the block climbs 4.0m up the ramp.
4)
a) When the plane is sitting on the runway, the bolts must supply an upwards force on the
engine that exactly cancels the downward force due to gravity. This force is:
Fg = mg = 1400kg ⋅ 9.8m/s 2 = 1.4 × 104 N
Since there are three bolts, and they share the load equally, each one must supply a force
of 4.6 x 103N.
b) Now the plane is accelerating upwards, and we’d like the engine to go with it. That
means a net upwards force must be applied to it, of magnitude:
F = ma = 1400kg ⋅ 2.60m/s = 3.64 × 103 N
This force is the sum of the upwards force applied by the bolts and the downward force
of gravity:
F = Fbolts + Fg
3.64 × 103 N = Fbolts + ( −1.4 × 104 N )
Fbolts = 1.8 × 104 N
Note that the force due to gravity is in the negative (downward) direction.
Each bolt supplies 1/3rd of the force, or 5.9 x 103N.
c) One of the goals of airplane design is to keep the mass as small as possible, meaning
that the minimal number of parts should be used. However, one can’t fasten the engine to
the plane with a single bolt, since the engine would be free to rotate around that bolt.
Two bolts would work, but if one happened to fail during flight, the engine would again
be free to rotate, and the plane might become impossible to control. With three bolts, any
one could fail, and the engine would remain securely fastened to the plane. So three bolts
is the minimum safe number to use.
9) This one is easier if we do part (c) first:
c) Each link in the chain is accelerating upwards at 2.50m/s2. Hence the net force on
each must be:
F = ma = ( 0.100kg ) ( 2.50m/s 2 ) j = 0.250Nj
a) To find the forces acting between the adjacent links, we start at the lowest one:
F12
Fg=mg
F12 is the force with which the second chain is pulling up the first one. Since we know
that the net force must be 0.250N upwards, we can find F12:
Fnet = 0.250Nj= ( F12 − Fg ) j
F12 = 0.250N + mg = 0.250N + ( 0.100kg ) ( 9.8m/s 2 )
= 1.23N
Now we look at the second link:
F23
F21
Fg=mg
There are two downward forces acting on this link – gravity and the force of the lowest
link both pull downwards. By Newton’s Third Law, though, we know that F21 has the
same magnitude as the force F12 acting on the lowest link. So we have:
Fnet = 0.250Nj = ( F23 − F12 − Fg ) j
F23 = 0.250N + 1.23N + 0.98N = 2.46N
The diagram for the third link would look just the same, so we have:
Fnet = 0.250Nj = ( F34 − F32 − Fg ) j
F34 = 0.250N + 2.46N + 0.98N = 3.69N
By doing the same steps one more time, we find the force between the fourth and fifth
links to be 4.92N.
b) The force FT pulling up on the top link must be enough to accelerate the entire chain
upwards at 2.50m/s2. To determine what force would do this, we don’t need to worry
about the individual links – we can treat the entire chain as a particle. We then have:
Fnet = FT − Fg = ( FT − mchain g ) j = mchain aj
FT = mchain ( a + g ) = ( 0.500kg ) ( 2.50m/s 2 + 9.80m/s 2 )
= 6.15N
10)
a) The total mass of the blocks is 4.4 kg. Since a net force of 3.2N is applied, the twoF 3.2N
block system must accelerate with a = =
= 0.91m/s 2 .
m 3.5kg
Now consider m2 by itself. It feels a force from block 1 which causes it to accelerate.
But since the two block remain in contact, we know that that both experience the same
acceleration. Thus the force of contact between the blocks must be:
F12 = m2a = (1.2kg ) ( 0.91m/s2 ) = 1.10N
b) If the force were applied from the other direction, the acceleration of the two-block
system would still be 0.73 m/s2 (in the opposite direction from that in part (a)). In this
case though, it is block 1 that feels only the force of contact between the two blocks.
Therefore this force must be:
F12 = m1a = ( 2.3kg ) ( 0.91m/s2 ) = 2.1N
While we may be surprised to find that the force of contact between the blocks depends
on the direction from which the external force is applied, we should note that the
configuration of masses is not left-right symmetric (i.e., the block on the left is more
massive). Thus we can’t assume that all properties of the system will remain the same
when the force is applied from the other direction.