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Physics 201 Homework 5
Feb 6, 2013
1. The (non-conservative) force propelling a 1500-kilogram car up a mountain
road does 4.70×106 joules of work on the car. The car starts from rest at sea level
and has a speed of 27.0 m/s at an altitude of 200 meters above sea level. Obtain
the work done on the car by the combined forces of friction and air resistance,
both of which are non-conservative forces.
-1.21 × 106 joules
Solution
The propelling force has the effect of increasing the energy level of the car. This
energy goes three places: kinetic energy (or speed), potential energy (or altitude),
and some is lost to the nonconservative friction. We know the energy going in
and we can calculate the amount that goes into kinetic and potential energies.
What is left must have been lost to the non-conservative forces. In symbols:
Wadded − Wlost = ∆E = ∆KE + ∆P E
We know the kinetic energy gained:
KE = 12 mv 2 = 12 (1500)(27.0)2 = 0.5468 × 106
And the potential energy gained:
P E = mgh = (1500)(9.80)(200) = 2.9400 × 106
Plugging these and the energy added we have:
(4.70 × 106 ) − Wlost = 0.5468 × 106 + 2.9400 × 106
=⇒ Wlost = 1.2133 × 106
It’s a technicality, but the question asks for the work done by the resistive forces.
Since we have calculated the work lost, we need to flip the sign to emphasize that
the energy level is decreasing as a consequence of the resistance.
2. A person pushes a 16.0-kilogram shopping cart at a constant velocity for a
distance of 22.0 meters. She pushes in a direction 29.0◦ below the horizontal.
A 48.0-newton frictional force opposes the motion of the cart. (a) What is the
magnitude of the force that the shopper exerts? Determine the work done by (b)
the pushing force, (c) the frictional force, and (d) the gravitational force.
Solution
(a) Since the cart is moving at constant velocity, the system is in equilibrium,
so the net force is zero. In this problem there are four forces involved. The
pushing force, the weight of the cart, the normal support force, and the kinetic
friction. The motion is horizontal, so we should align the coordinate that way.
The components of the net force are:
Fnet,x = Px − F
Fnet,y = Py − W + N
Both of these components sum to zero. Since we are given the friction force
(48.0 newtons), we know that the x-component of the pushing force is also 48.0
newtons. Since we know that the canonical angle of the push is 331◦ , we have:
Px = P cos 331◦ = 48.0
=⇒ P = 54.881
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(a) 54.9 newtons
(b) 1060 joules
(c) -1060 joules
(d) none
(b) The work done by any force is the product of the displacement and the
component of the force parallel to the displacement. In this case, x = 22.0 meters
and Px = 48.0 newtons. Thus,
W = xPx = (22.0)(48.0) = 1056
(c) Similarly for the friction, but this force opposes the motion so its component
is negative:
W = xFx = (22.0)(−48.0) = −1056
(d) The gravitational force has no component in the x direction. Thus,
W = xWx = (22.0)(0) = 0
3. A 0.075-kilogram arrow is fired horizontally. The bowstring exerts an average
force of 65 newtons on the arrow over a distance of 0.90 meters. With what speed
does the arrow leave the bow?
39 m/s
Solution
The work done by the bowstring is
W = xFx = (0.90)(65) = 58.5
This work has the effect of increasing the energy of the arrow. In this case it
increases the kinetic energy of the arrow. The initial kinetic energy is zero, so
KEf = 58.5
Using the formula for kinetic energy, KE = 12 mv 2 , we can derive the final resulting
speed:
2
1
2 (0.075)(v) = 58.5 =⇒ v = 39.497
4. An extreme skier, starting from rest, coasts down a mountain slope that makes
an angle of 25.0◦ with the horizontal. The coefficient of kinetic friction between
her skis and the snow is 0.200. She coasts down a distance of 10.4 meters before
coming to the edge of a cliff. Without slowing down, she skis off the cliff and
lands downhill at a point whose vertical distance is 3.50 meters below the edge.
How fast is she going just before she lands?
Solution
There are two phases to the motion in this problem. The first coasting phase, is
basically an inclined plane problem from Chapter 4. At the end of this phase the
skier will have a certain velocity angled down 25◦ from the horizontal. The second
phase is standard projectile problem which we could solve using the techniques
from Chapter 3. However, this since we are only asked about the speed, we could
use kinetic energy to answer the question. Either approach will work. But since
this is a Chapter 6 problem, it seems appropriate to use energy.
In the first phase, there are three forces at work: weight, support, and friction.
Using coordinates aligned with the motion (assume the slope is to the right), the
canonical angles associated with these forces are 295◦ , 90◦ , and 180◦ , respectively.
The magnitudes of these forces are W = mg, N , and Fk = (0.200)(N ), respectively. We don’t know m, but we will assume it cancels out. We don’t know N ,
but it must counter-balance the y-component of the weight. Thus, N + Wy = 0.
Therefore,
N = −Wy = −W sin θ = −(9.80)(m) sin 295◦ = (8.8818)(m)
Now there is motion in the x-direction. According to Newton’s second law we
have Fnet = ma. The net force is in the x-direction, but friction points in the
negative direction, so
Fnet,x = Wx − Fk = (m)(9.80) cos 295◦ − (0.200)(8.8818)(m) = (2.3653)(m)
2
10.9 m/s
So, the acceleration of the skier down the slope is 2.3653 m/s2 .
Now we have a constant acceleration problem. The quantities are
a = 2.3653
v0 = 0
v=?
x = 10.4
The equation to use is v 2 = v02 + 2ax. Thus,
(v)2 = (0)2 + (2)(2.3653)(10.4) =⇒ v = 7.0141
This is the endpoint of the first phase of motion, but it is also the beginning of
the second phase. We decided to use energy considerations to work this phase.
We know that she is 3.50 meters above her landing point. This means she has a
certain amout of potential energy:
P E = mgh = (m)(9.80)(3.50) = (34.300)(m)
And her initial kinetic energy is
KE = 12 mv 2 = 12 (m)(7.0141)2 = (24.599)(m)
Since no energy is lost, the potential energy is converted into additional kinetic
energy at the end of the drop. So the final kinetic energy is
KE = (34.300)(m) + (24.599)(m) = (58.899)(m)
Using the formula for kinetic energy, KE = 21 mv 2 , we can determine her final
speed:
(58.899)(m) = 21 (m)(v)2 =⇒ v = 10.853
5. “Rocket man” has a propulsion unit strapped to his back. He starts from rest
on the ground, fires the unit, and is propelled straight upward. At a height of
16 meters, his speed is 5.0 m/s. His mass, including the propulsion unit, has the
approximately constant value of 136 kilograms. Find the work done by the force
generated by the propulsion unit.
23 kJ
Solution
The gravitational force is conservative, so we can say that
W = ∆E = ∆KE + ∆P E
Now, initially “Rocket man” has no kinetic or potential energy because he is at
rest and on the ground. But at the end point, he has both kinetic and potential
energy. Thus,
KE = 12 mv 2 = 12 (136)(5.0)2 = 1700
and
P E = mgh = (136)(9.80)(16) = 21325
Now, his initial mechanical energy is zero. But the final mechanical energy is
E = KE + P E = 1700 + 21325 = 23025
The work done by the propulsion unit is where this energy came from.
6. In 2.0 minutes, a ski lift raises four skiers at constant speed to a height of
140 meters. The average mass of each skier is 65 kilograms. What is the average
power provided by the tension in the cable pulling the lift?
Solution
3
3000 watts
Since the work done by the lift has the effect of increasing the potential energy
of the skiers, we know by the work-energy theorem that
W = ∆P E = mgh = (4 × 65)(9.80)(140) = 356720
Since this quantity of work is done over a 2.0 minute time period (or 120 seconds),
by definition, the power involved is
P =
356720
W ork
=
= 2972.7
∆t
120
7. A small lead ball, attached to a 1.5 meter rope, is being whirled in a circle
that lies in the vertical plane. The ball is whirled at a constant rate of three
revolutions per second and is released on the upward part of the circular motion
when it is 0.75 meters above the ground. The fall travels straight upward. In the
absence of air resistance, to what maximum height above the ground does the
ball rise?
42 meters
Solution
At the top of its trajectory, the kinetic energy is momentarily zero—all the energy
is in the gravitational potential energy mgh. From this we can calculate the
height. Since we know the initial height, all we need to determine is the initial
speed.
The ball moves in uniform circular motion with a radius of 1.5 meters. Each
revolution is 2πr = 9.4248 meters. The speed is three times this value per second:
v = (3)(2π)(1.5) = 28.274
So, the energy of the ball is
KE = 21 mv 2 = 12 (m)(28.274)2 = 399.72
PE = mgh = (m)(9.8)(0.75) = 7.35
For a total of (407.07)(m) joules of energy. Therefore the maximum height must
be
PE = mgh = (m)(9.8)(h) = (407.07)(m) =⇒ h = 41.538
8. A pendulum consists of a small object hanging from the ceiling at the end
of a string of negligible mass. The string has a length of 0.75 meters. With the
string hanging vertically, the object is given an initial velocity of 2.0 m/s parallel
to the ground and swings upward on a circular arc. Eventually, the object comes
to a momentary halt at a point where the string makes an angle θ with its initial
vertical orientation and then swings back downward. Find the angle θ.
Solution
Initially the height of the pendulum is zero, so its potential energy is zero. The
kinetic energy is
KE = 21 mv 2 = 12 (m)(2.0)2 = (2.0)(m)
We can use this to determine the height the pendulum rises, since the energy will
all be converted to potential:
PE = mgh = (m)(9.8)(h) = (2.0)(m) =⇒ h = 0.20408
From this we should be able to calculate the angle in question using some trig. I
will help to draw a little picture here:
4
43◦
θ
0.75
0.75
0.20
Figure 1: Problem 6.46
From the picture we can see the triangle defined by the angle in which we are
interested. We know the hypotenuse is 0.75 meters and the adjacent side is (0.75
− 0.20408) = 0.54592. These sides are connected via the cosine function. Thus:
cos θ =
0.54592
=⇒ θ = 43.290
0.75
◦
9. A semitrailer is coasting downhill along a mountain highway when its brakes
fail. The driver pulls onto a runaway-truck ramp that is inclined at an angle of
14.0◦ above the horizontal. The semitrailer coasts to a stop after traveling 154
meters along the ramp. What was the truck’s initial speed? Neglect air resistance
and friction
27.0 m/s
Solution
We are asked about the initial state, so let’s start with the end state instead. We
know the semi comes to a stop, so there is no kinetic energy. It is at a certain
height, which is potential energy. But how high? The height is actually the
opposite side of the triangle formed by the ramp. The hypotenuse is 154 meters
and the angle is 14.0◦ . Thus:
sin 14.0
◦
= h 154 =⇒ h=37.256 So, the potential energy is
PE = mgh = (m)(9.8)(37.256) = (365.11)(m)
This is how much kinetic energy it has when it hits the ramp. Thus,
(365.11)(m) = 21 mv 2 =⇒ v = 27.023
10. A 1300-kilogram car is to accelerate from rest to a speed of 30.0 m/s in a
time of 12.0 seconds as it climbs a 15◦ hill. Assuming uniform acceleration, what
minimum horsepower is needed to accelerate the car in this way?
Solution
We could use the techniques from Chapter 4 to solve this, but using energy is
much easier. The initial energy of the car is zero: it starts at rest (no kinetic)
and is at the bottom of the hill (no potential). As the engine works on the car, it
gains energy. At the top of the hill it has kinetic energy:
KE = 12 mv 2 = 12 (1300)(30.0)2 = 5.85 × 105
Its potential energy is governed by it height. In order to figure this out we need
to determine how far it travels. We can use
x = 21 (v + v0 )(t) = 12 (30.0 + 0)(12.0) = 180
5
132 horsepower
Since the hill is at a 15◦ incline, the total elevation change is
h = x sin θ = (180) sin 15◦ = 46.587
So the car’s final potential energy is
P E = mgh = (1300)(9.8)(46.587) = 5.9352 × 105
The work done by the engine is the combination of the kinetic and potential
energies:
W = 1.1785 × 106
Finally, power is defined as work done per second, so:
P =
W
1.1785 × 106
=
= 98210
t
12.0
Since there are 746 watts in one horsepower, we have
P = 98210/746 = 131.65 hp
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