Download A 75.0-kg skier starts from rest and slides down a

PHYS 201
Exercise Set #17
Solutions
1
Problem 6.22 (8 points) A 75.0-kg skier starts from rest and slides down a 32.0-m frictionless slope that is inclined at an angle of 15.0◦ with the horizontal. Ignore air resistance.
(a) Calculate the work done by gravity on the skier and the work done by the normal
force on the skier. (b) If the slope is not frictionless so that the skier has a final velocity
of 10.0 m/s, calculate the work done by gravity, the work done by the normal force, the
work done by friction, the force of friction (assume it was constant), and the coefficient
of kinetic friction.
(a) The drop in the skier’s height is ∆h = (32.0 m)(sin 15.0◦ ) = 8.28 m. The work
done by gravity is positive, because the skier’s height is changing in the same direction
that gravity is pulling. The work is
Wg = mg∆h = (75.0 kg)(9.8 m/s2 )(8.28 m) = 6090 J
The work done by the normal force is WN = 0 since the normal force is at 90◦ to the
displacement. There is no component of that force in the direction of displacement, so
it cannot do any work.
(b) The work done by gravity is just the same:
Wg = 6090 J
Again, the work done by the normal force is zero: WN = 0 . To find the work done
by friction, we use the Work-Kinetic Energy Principle:
Wnet = ∆K
1 2 1 2
Wg + Wf =
mv − mv
2 f 2 i
1
(75.0 kg)(10.0 m/s)2 − 0
6090 J + Wf =
2
Wf = 3750 J - 6090 J = -2340 J
We can find the force of friction by the definition of work:
Wf = fk d
-2340 J
Wf
=
fk =
d
32.0 m
fk = 73 N, uphill
Finally, to find the coefficient of kinetic friction, we must know the normal force. The
free-body diagram is the same as just about any object moving along an incline. The
normal force is equal and opposite to the component of weight (mg) perpendicular to
the surface. We have seen over and over that this is
N = mg cos 15◦ = (75 kg)(9.8 m/s2 )(0.966) = 710 N
PHYS 201
Exercise Set #17
2
Solutions
Thus,
fk = µk N
73 N = µk (710 N)
73 N
µk =
= 0.10
710 N
Problem 6.26 (6 points) A cart moving to the right passes point 1 in Fig. 6.30 at a speed
of 20.0 m/s. Let g = 9.81 m/s2 . (a) What is the speed of the cart as it passes point
3? (b) Will the cart reach position 4? Neglect friction.
4
2
3
15.0 m
1
20.0 m
10.0 m
(a) If we can ignore friction, then the total mechanical energy remains the same.
KEi + PEi = KEf + PEf
1 2
1 2
mvo + 0 =
mv + mgh
2
2
1
1 2
(20.0 m/s)2 =
v + (9.81 m/s2 )(10.0 m)
2
2
v 2 = 203.8 m2 /s2
v =
14.3 m/s
(b) We can see if we get a sensible answer, using the same method as above:
1 2
1 2
mvo + 0 =
mv + mgh
2
2
1
1 2
(20.0 m/s)2 =
v + (9.81 m/s2 )(20.0 m)
2
2
v 2 = 7.6 m2 /s2
v = 2.8 m/s
Yes, it will make it to position 4. If we had obtained a negative value for v 2 , we would
know that the total mechanical energy was not enough for the cart to reach position
4.
PHYS 201
Exercise Set #17
Solutions
3
Problem 6.27 (6 points) A cart starts from position 4 in Fig. 6.30 with a velocity of 15
m/s to the left. Find the speed with which the cart reaches positions 1, 2, and 3.
Neglect friction.
4
2
3
15.0 m
20.0 m
10.0 m
1
We get to ignore friction, so the total mechanical energy is conserved. We will take
the height at position 1 to be our reference for potential energy: PE = 0 at position 1.
We are given the velocity at position 4, so we can find both the kinetic and potential
energies at position 4.
Therefore the total mechanical energy of the system is
1 2
1
mv + mgh =
m(15 m/s)2 + m(9.8 m/s2 )(20.0 m)
2
2
= m(308.5 m2 /(s2 -kg))
At Position 1 h = 0, so
1 2
mv + mgh1 = m(308.5 m2 /(s2 -kg))
2 1
v12 = 2(308.5 m2 /s2 )
v1 =
At Position 2 h = 15.0 m, so
1 2
mv + mgh3
2 2
1 2
v
2 2
1 2
v
2 2
v22
= m(308.5 m2 /(s2 -kg))
+ (9.8 m/s2 )(15.0 m) + 308.5 m2 /s2
= 308.5 m2 /s2 − 147 m2 /s2
= 2(161.5 m2 /s2 )
v2 =
At Position 3 h = 10.0 m, so
1 2
mv + mgh2
2 3
1 2
v
2 3
1 2
v
2 3
v32
24.8 m/s at position 1
18.0 m/s at position 2
= m(308.5 m2 /(s2 -kg))
+ (9.8 m/s2 )(10.0 m) + 308.5 m2 /s2
= 308.5 m2 /s2 − 98 m2 /s2
= 2(210.5 m2 /s2 )
v3 =
20.5 m/s at position 3
PHYS 201
Exercise Set #17
Solutions
4
L cosθ
Problem 6.28 (6 points) Bruce stands on a bank beside a pond, grasps the end of a 20.0-m
long rope attached to a nearby tree and swings out to drop into the water. If the rope
starts at an angle of 35◦ with the vertical, what is Bruce’s speed at the bottom of the
swing.
Looking at the diagram, Bruce drops vertically a distance h. His decrease in potential energy is then mgh. This will “show
up” as an increase in kinetic energy. More
formally,
L
θ
KEi + PEi = KEf + PEf
1 2
mv + 0
0 + mgh =
2
where we have chosen his ending height as
our “zero” level for potential energy calculations. It is clear from the diagram that
h
h = L − L cos θ = L(1 − cos θ)
so
h = (20 m)(1 − cos 35◦ ) = 3.62 m
Thus,
1 2
mv
2
= 2gh = 2(9.81 m/s2 )(3.617 m)
mgh =
v2
v =
8.4 m/s