Download 6.6, Quiz

Physics 2001
Problem Set 6 Solutions
Jeff Kissel
October 23, 2006
1. You are moving out of town and need to load a heavy box of
mass m into the back of the moving truck. You can either pick
the box straight up and set it in the truck (a height h above the
ground) or slide the box a distance L up a ramp into the back
of the truck. Assume the ramp has good rollers on it, so that it
is frictionless.
(a) Which way of loading the box is easier (meaning which way
requires less force?)
(b) Which way requires more work? (Calculate the work done
in each case to find out.)
Figure 1: Left: A free-body diagram of lifting the box straight up onto the
truck. Right: A free-body diagram of pushing the box on a frictionless ramp.
(a)Which takes less force? Just from real life experience, we know that
rolling the box up the frictionless ramp will require less force, but let’s
prove it to ourselves using Newton’s equations.
When simply lifting the box (as in the left-hand side of Figure 1), the only
forces are in the ŷ direction. The force you apply must be greater than
the force of gravity in order for it to move, but we’re lazy (or at least I
1
Jeff Kissel, October 23, 2006
Physics 2001, Problem Set 6 Solutions
am) so let’s consider the least amount of force needed to lift the box. This
force will be just a little bit more than when the force is in equilibrium
with gravity, or
X
Fy
0
Flif ting
= Flif ting − mg
= mg
(1)
When rolling up the ramp, we only have to apply the force in the (tilted)
x̂ direction (shown in the right-hand side of Figure 1). The minimum force
in the direction now needs only compensate the x̂ component of gravity,
or
X
Fx
0 =
Framping =
Framping − mg sin θ
mg sin θ
(2)
The only difference between Eq. 1 and 2 is the factor of sin θ. We know
the ramp will be at some angle less than θ = 90◦ , so sin θ will be less than
one, and Framping is less than Flif ting as we expected.
(b)What takes less work? We will use the general definition of work which
states that
≡
W
F~ · d~
(3)
where F~ and d~ are the force and displacement vectors, and the “ · ” is called
the “dot product” or “scalar product.” The dot product is different from
regular multiplication, because it deals with two vectors, which a direction
associated with them in addition to a magnitude. The dot product is
defined as
~·B
~ ≡
A
or =
~ B|
~ cos φ
|A||
AB cos φ
(4)
~ = A, and |B|
~ = B are the magnitudes of each vector, and φ is
where |A|
the angle between the vectors.
As stated in the first part of the answer, the only forces we need to worry
about are the force we apply, F~ and that of gravity F~ = m~g (because in
each case we’re worried only about the work that we have to do). So, for
~ = h, and |Fg | = m|~g | = mg, and the vectors are 180◦ from each
lifting, |d|
2
Jeff Kissel, October 23, 2006
Physics 2001, Problem Set 6 Solutions
other, so the work done by gravity is
Wg
=
=
Fg · d~
mg h cos (180◦ )
Wg
=
−mgh
(5)
To find the work we have to do WL , we must use the kinetic energy
theorem: the change in kinetic energy is equal to the total work.
∆KE = Wtot
(6)
This is useful, because we’re lifting the box from rest on the ground to
rest on the truck, so vi = vf = 0. The change in kinetic energy is then
zero, since KE ≡ 12 mv 2 . So, since the work we do is the only other work
on the system,
0 =
WL
WL
=
=
WL + Wg
−Wg
mgh
(7)
Now for pushing up the ramp. As before we’re only concerned with work
we do, and the work done gravity, Wg,x . The magnitude of the gravita~ = L, and the angle between
tional force is again |Fg | = mg, but now |d|
◦
the vectors is φ = θ + 90 . The work done by gravity is then
Wg
Wg
=
Fg · ~g
=
=
mg L cos φ
mgL cos (θ + 90◦ )
=
cos (θ + 90◦ ) = sin θ
mgL sin θ
=
mgh
(8)
Huh! Apparently, because of a few trig tricks, the work done by ramping
is exactly the same as lifting. It only feels like less force, because the
normal force of the ramp is helping you lift the box.
2. A standard size Snickers candy bar contains 280 Calories, which
is equivalent to 1.2 × 106 J of energy. Assume a 65 kg mountain
climber can convert all of this energy into work he does climbing
a mountain. How many Snickers bars would he have to eat to
climb Mt. Everest (29, 028 f t. = 5.5 mi. = 8, 848m)?
As we found out in the previous problem, it takes to the same amount of
work to travel up the side of the mountain as it does to take an elevator
3
Jeff Kissel, October 23, 2006
Physics 2001, Problem Set 6 Solutions
from the heart of the mountain to the top. Let’s do the latter, since the
calculation is more straight forward.
The work done by gravity as the climber moves from the bottom to the
top will be just just like lifting the box,
Wg
=
=
F~g · d~
mg h cos (180◦ )
Wg
=
−mgh
(9)
which means that work that the climber has to do is the exact opposite
since there’s no other forces, and he’ll move from rest to rest,
:0
∆KE
= Wtot
0 = Wclimber + Wg
Wclimber
= mgh
= (65 kg)(9.8 m s−2 )(8, 848 m)
(10)
Wclimber
= 5.636 × 106 J
(11)
We’ve already established that work has units of energy (joules, J) when
we set the change in kinetic energy equal to the work. So, all we need to
do is convert joules to candy bars, my favorite unit of energy!
Wclimber
=
(5.636 × 106 J) ×
Wclimber
=
4.69 candy bars
1 candy bar
1.2 × 106 J
(12)
Yes Dr. Huge. There is a lot of energy in food.
3. A ball on a frictionless track is released from rest at a point A.
What is the speed of the ball at point C? At point D?
One of the big concepts in physics (right up there with F = ma) is conservation of energy. As long as the system is conservative (i.e. energy is not
lost to friction, heat, sound, or otherwise) the sum of the kinetic energy
and the potential energies is equal to the same sum after some action. In
equation form, that’s
=
=
Ei
KEi + P Ei
Ef
KEf + P Ef
(13)
The kinetic energy is defined as
KE =
4
1
mv 2
2
(14)
Jeff Kissel, October 23, 2006
Physics 2001, Problem Set 6 Solutions
Figure 2: A ball rolling down a frictionless track.
Since there is no friction, the only source of potential energy on the track
is the gravitation potential on the ball. The gravitational potential energy
is equivalent to the work done by gravity, defined as
W
PE
=
=
PE
=
F~ · d~
(m~g) · d~
~
m (~g · d)
=
~ cos θ)
m(|~g||d|
~ =h
But d~ points in the same direction as ~g , so cos θ = cos (0◦ ) = 1. |d|
and |~g| = g, which means the potential energy is just
P E = mgh
(15)
With these two definitions (Eq. 14 and 15), and Eq. 13, we can arrive at
the answer for the velocity at point C. Note that the problem states that
hi = h, hf = 0.35h, and the ball is released from rest, so vi = 0.
KEi + P Ei
0
1 2
mvi + mghi
2
1
mv 2
2 f
vf2
=
KEf + P Ef
=
1
mv 2 + mghf
2 f
= mghi − mghf
=
=
2g(hi − hf )
2g(h − 0.35h)
=
2g(0.65h)
p
1.3gh
vC =
=
vf
(16)
(17)
As for the speed at point D, it’s virtually the same steps, but at Eq. 16
5
Jeff Kissel, October 23, 2006
Physics 2001, Problem Set 6 Solutions
we just set hf = h, as the Figure 2 indicates. This means
vf2
= 2g(hi − hf )
= 2g(h − h)
vf
= vD = 0
(18)
4. Quiz: What is the gravitational potential energy of a penny
(m = 0.0025 kg) on top of the Sears’ Tower - a height of 442 m?
The gravitational potential energy, as we’ve discussed, is
PE
~
= m (~g · d)
~ cos θ
= m|~g ||d|
:1
◦
(0
= m g h
cos
)
= mgh
PE
= (0.0025 kg)(9.8 m s
= 10.829 J
6
(19)
−2
)(442 m)
(20)