Download Homework 6 - NMSU Physics

Solution to HW#6 – Phys 211 – Fall 2002
8.
REASONING AND SOLUTION The speed of a particle doubles and then doubles
again, because a net external force acts on it. Let m and v0 represent the mass and
initial speed of the particle, respectively. During the first doubling, the change in the
kinetic energy of the particle is
KE f − KE 0 = 2 m ( 2v0 ) − 2 mv0 = 2 mv0
2
1
1
3
2
2
During the second doubling, the change in the kinetic energy of the particle is
KE f − KE 0 = 2 m ( 4v0 ) − 2 m ( 2v0 ) = 6mv0
1
2
1
2
2
From the work-energy theorem, we know that a change in kinetic energy is equal to
the work done by the net external force. Therefore, more work is done by the net
force during the second doubling.
15. REASONING AND SOLUTION Car A turns off its engine and coasts up the hill.
Car B keeps its engine running and drives up the hill at constant speed. If air
resistance and friction are negligible, then only the motion of car A is an example of
the principle of conservation of mechanical energy. During the motion of car A,
there are no nonconservative forces present. As car A coasts up the hill it loses
kinetic energy and gains an equal amount of gravitational potential energy. The
motion of car B is not described by the principle of conservation of mechanical
energy, because the driving force that propels car B up the hill is a nonconservative
force.
6.P.6. REASONING AND SOLUTION
a. In both cases, the lift force L is perpendicular to the displacement of the plane,
and, therefore, does no work. As shown in the drawing in the text, when the plane is in
the dive, there is a component of the weight W that points in the direction of the
displacement of the plane. When the plane is climbing, there is a component of the
weight that points opposite to the displacement of the plane. Thus, since the thrust T is
the same for both cases, the net force in the direction of the displacement is greater for
the case where the plane is diving. Since the displacement s is the same in both cases,
more net work is done during the dive .
b. The work done during the dive is Wdive = (T + W cos 75°) s , while the work
done during the climb is Wclimb = (T + W cos 115°) s . Therefore, the difference between
the net work done during the dive and the climb is
Wdive – Wclimb = (T + W cos 75°) s – (T + W cos 115°) s = Ws (cos 75° – cos 115°)
= (5.9 × 10 4 N)(1.7 × 103 m)(cos 75° – cos 115°) = 6.8 × 10 7 J
6.P.9. SSM REASONING AND SOLUTION According to Equation 6.1, the work
done by the husband and wife are, respectively,
[ Husband ]
WH = ( FH cos θ H )s
[ Wife]
WW = ( FW cos θ W )s
Since both the husband and the wife do the same amount of work,
( FH cos θ H )s = ( FW cos θ W )s
Since the displacement has the same magnitude s in both cases, the magnitude of the
force exerted by the wife is
FW = FH
cos θ H
cos 58°
= (67 N)
= 45 N
cos θ W
cos 38°
6.P.16. REASONING AND SOLUTION From the work-energy theorem, Equation 6.3,
1
1
1
W = mvf2 − mv02 = m vf2 − v02 
2
2
2 

a.
W = (7420 kg)  (8450 m/s) 2 – (2820 m/s)2  =
2


1
2.35 × 1011 J
1
W = (7420 kg) (2820 m/s)2 − (8450 m/s)2  = –2.35 × 1011 J
2


6.P.21. REASONING It is useful to divide this problem into two parts. The first part
involves the skier moving on the snow. We can use the work-energy theorem to find her
speed when she comes to the edge of the cliff. In the second part she leaves the snow and
falls freely toward the ground. We can again employ the work-energy theorem to find her
speed just before she lands.
b.
SOLUTION The drawing at the right shows the
three forces that act on the skier as she glides on the snow.
The forces are: her weight mg, the normal force FN, and
the kinetic frictional force fk. Her displacement is labeled
as s. The work-energy theorem, Equation 6.3, is
FN
+y
fk
s
65.0°
25.0°
mg
W = 12 mvf2 − 12 mv02
where W is the work done by the net external force that acts on the skier. The work
done by each force is given by Equation 6.1, W = ( F cos θ ) s , so the work-energy
theorem becomes
( mg cos 65.0 ) s + ( f k cos 180 ) s + ( FN cos 90 ) s
= 12 mvf2 − 12 mv02
W
Since cos 90° = 0, the third term on the left side can be eliminated. The magnitude
fk of the kinetic frictional force is given by Equation 4.8 as f k = µ k FN . The magnitude
FN of the normal force can be determined by noting that the skier does not leave the
surface of the slope, so ay = 0 m/s2. Thus, we have that ΣFy = 0, so
FN − mg cos 25.0° = 0
or
FN = mg cos 25.0°
Σ Fy
The magnitude of the kinetic frictional force becomes f k = µ k FN = µ k mg cos 25.0 .
Substituting this result into the work-energy theorem, we find that
( mg cos 65.0 ) s + ( µk mg cos 25.0 ) ( cos 180 ) s
= 12 mvf2 − 12 mv02
W
Algebraically eliminating the mass m of the skier from every term, setting cos
180° = –1 and
v0 = 0 m/s, and solving for the final speed vf, gives
vf = 2 gs ( cos 65.0 − µ k cos 25.0
)
= 2 ( 9.80 m/s 2 ) (10.4 m ) cos 65.0 − ( 0.200 ) cos 25.0  = 7.01 m/s
The drawing at the right shows her displacement s during free
fall. Note that the displacement is a vector that starts where she
leaves the slope and ends where she touches the ground. The only
force acting on her during the free fall is her weight mg. The workenergy theorem, Equation 6.3, is
3.50 m
W = 12 mvf2 − 12 mv02
The work W is that done by her weight, so the work-energy theorem becomes
θ
s
mg
( mg cosθ ) s
= 12 mvf2 − 12 mv02
W
In this expression θ is the angle between her weight (which points vertically
downward) and her displacement. Note from the drawing that s cos θ = 3.50 m.
Algebraically eliminating the mass m of the skier from every term in the equation above
and solving for the final speed vf gives
vf = v02 + 2 g ( s cos θ )
=
( 7.01 m/s )
2
+ 2 ( 9.80 m/s 2 ) ( 3.50 m ) = 10.9 m/s
6.P.28. REASONING AND SOLUTION
a. From the definition of work, W = Fs cos θ. For upward motion θ = 180° so
W = – mgs = – (71.1 N)(2.13 m – 1.52 m) = –43 J
b. The change in potential energy is
∆PE = mghf – mgh0 = (71.1 N)(2.13 m – 1.52 m) = +43 J
6.P.36. REASONING Since air resistance is being neglected, the only force that acts on
the golf ball is the conservative gravitational force (its weight). Since the maximum
height of the trajectory and the initial speed of the ball are known, the conservation of
mechanical energy can be used to find the kinetic energy of the ball at the top of the
highest point. The conservation of mechanical energy can also be used to find the speed
of the ball when it is 8.0 m below its highest point.
a.
SOLUTION
The conservation of mechanical energy, Equation 6.9b, states that
1 mv 2
f
2
+ mghf = 12 mv02 + mgh0
KEf
Solving this equation for the final kinetic energy, KEf, yields
KE f = 12 mv02 + mg ( h0 − hf )
=
b.
1
2
( 0.0470 kg )( 52.0 m / s )2 + ( 0.0470 kg ) ( 9.80 m / s 2 ) ( 0 m − 24.6 m ) =
The conservation of mechanical energy, Equation 6.9b, states that
52.2 J
1 mv 2
f
2
+ mghf =
1 mv 2
0
2
Ef
+ mgh0
Ef
The mass m can be eliminated algebraically from this equation, since it appears as a
factor in every term. Solving for vf, and noting that the final height of the ball is hf =
24.6 m – 8.0 m = 16.6 m, we have that
vf =
v02 + 2 g ( h0 − hf ) =
( 52.0 m / s )2 + 2 ( 9.80 m / s2 ) ( 0 m − 16.6 m ) =
48.8 m / s
6.P.41. SSM REASONING Friction and air resistance are being ignored. The normal
force from the slide is perpendicular to the motion, so it does no work. Thus, no net work
is done by nonconservative forces, and the principle of conservation of mechanical
energy applies.
SOLUTION Applying the principle of conservation of mechanical energy to the
swimmer at the top and the bottom of the slide, we have
1
1
2
2
mvf + mghf = mv0 + mgh0
2
2
Ef
E0
If we let h be the height of the bottom of the slide above the water, hf = h , and
h0 = H . Since the swimmer starts from rest, v0 = 0 m/s, and the above expression
becomes
1 2
v
2 f
Solving for H, we obtain
+ gh = gH
v2
H = h+ f
2g
Before we can calculate H, we must find vf and h. Since the velocity in the
horizontal direction is constant,
∆x 5.00 m
vf =
=
= 10.0 m/s
∆t 0.500 s
The vertical displacement of the swimmer after leaving the slide is, from Equation
3.5b (with down being negative),
y = 2 ay t 2 = 2 (–9.80 m/s 2 )(0.500 s)2 = −1.23 m
1
1
Therefore, h = 1.23 m. Using these values of vf and h in the above expression for
H, we find
v2
(10.0 m/s)2
H = h + f = 1.23 m +
2 = 6.33 m
2g
2(9.80 m/s )
6.P.45. REASONING AND SOLUTION The work-energy theorem can be used to
determine the change in kinetic energy of the car according to Equation 6.8:
Wnc =
(
1
2
mvf2 + mghf
)−(
Ef
1
2
mv02 + mgh0
)
E0
The nonconservative forces are the forces of friction and the force due to the
chain mechanism. Thus, we can rewrite Equation 6.8 as
Wfriction + Wchain = ( KE f + mghf ) − ( KE 0 + mgh0 )
We will measure the heights from ground level. Solving for the change in kinetic
energy of the car, we have
KE f − KE 0 = Wfriction + Wchain − mghf + mgh0
= − 2.00×104 J + 3.00 × 104 J − (375 kg)(9.80 m/s 2 )(20.0 m)
+ (375 kg)(9.80 m/s2 )(5.00 m) = –4.51×104 J
6.P.49. REASONING AND SOLUTION The force exerted by the bat on the ball is the
only non-conservative force acting. The work due to this force is
Wnc = mvf – mv0 + mg ( hf – h0 )
1
2
2
1
2
2
Taking h0 = 0 m at the level of the bat, v0 = 40.0 m/s just before the bat strikes the
ball and vf to be the speed of the ball at hf = 25.0 m, we have
vf =
=
2Wnc
m
+ v02 – 2 ghf
2(70.0 J)
2
+ ( 40.0 m/s ) –2(9.80 m/s 2 )(25.0 m) = 45.9 m/s
0.140 kg