arellano (aa39398) – Practice Homework Chapter 5 Solutions – Weathers – (17101) 1
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001 (part 1 of 8) 10.0 points
Assume you are on a planet similar to Earth
where the acceleration of gravity is approximately 10 m/s
2
and the positive directions for
displacement, velocity, and acceleration are
upward.
At time t = 0 s, an elevator is at a displacement of x = 0 m with a velocity of v = 0 m/s.
A student whose normal weight is 400 N
stands on a scale in an elevator and records
the scale reading as a function of time. The
data are shown in the graph.
0 5 10 15 20
0
200
400
600
800
time (s)
Scale Reading (N)
In which direction does the force due to the
scale act on the student?
1. backward
2. right
3. up correct
4. The net force on the student is zero, so
the force the scale exerts on the student has
no direction.
5. forward
6. down
7. left
Explanation:
The scale exerts a normal force (Scale Reading Fs), opposite to the force of gravity m g;
i.e., upward.
002 (part 2 of 8) 10.0 points
What is the mass of the student?
1. m ≈ 70 kg
2. m ≈ 30 kg
3. m ≈ 10 kg
4. m ≈ 60 kg
5. m ≈ 40 kg correct
6. m ≈ 80 kg
7. m ≈ 50 kg
8. m ≈ 90 kg
9. m ≈ 20 kg
Explanation:
Since W = m g, we have
m =
W
g
=
400 N
10 m/s
2
= 40 kg .
003 (part 3 of 8) 10.0 points
Calculate the acceleration of the elevator during the fourth 5 s interval (from 15 s to 20
s).
Correct answer: 5 m/s
2
.
Explanation:
The net force Fnet on the student is
Fnet = Fs − 400 N = m a ,
so
a =
Fs − 400 N
m
.
The accelerations during the 5-second intervals are
a0→5 =
400 N − 400 N
40 kg = 0 m/s
2
arellano (aa39398) – Practice Homework Chapter 5 Solutions – Weathers – (17101) 2
a5→10 =
500 N − 400 N
40 kg = 2.5 m/s
2
a10→15 =
400 N − 400 N
40 kg = 0 m/s
2
a15→20 =
600 N − 400 N
40 kg = 5 m/s
2
.
004 (part 4 of 8) 10.0 points
Choose the correct plot for acceleration vs
time.
1.
0 5 10 15 20
−10
−5
0
5
10
time (s)
Acceleration (m/s
2
)
2.
0 5 10 15 20
−10
−5
0
5
10
time (s)
Acceleration (m/s
2
)
correct
3.
0 5 10 15 20
−10
−5
0
5
10
time (s)
Acceleration (m/s
2
)
4.
0 5 10 15 20
−10
−5
0
5
10
time (s)
Acceleration (m/s
2
)
5.
0 5 10 15 20
−10
−5
0
5
10
time (s)
Acceleration (m/s
2
)
6.
0 5 10 15 20
−10
−5
0
5
10
time (s)
Acceleration (m/s
2
)
7.
0 5 10 15 20
−10
−5
0
5
10
time (s)
Acceleration (m/s
2
)
8.
0 5 10 15 20
−10
−5
0
5
10
time (s)
Acceleration (m/s
2
)
Explanation:
Use the results calculated in Part 2 to construct the acceleration graph:
0 5 10 15 20
−10
−5
0
5
10
time (s)
Acceleration (m/s
2
)
arellano (aa39398) – Practice Homework Chapter 5 Solutions – Weathers – (17101) 3
005 (part 5 of 8) 10.0 points
What is the velocity of the elevator at the end
of the fourth 5 s interval (at 20 s)?
Correct answer: 37.5 m/s.
Explanation:
v = v0 + a t
At the end of five second intervals, we have
v1 = v0 + a1 t
= 0 m/s + (0 m/s
2
) (5 s)
= 0 m/s
v2 = v1 + a2 t
= 0 m/s + (2.5 m/s
2
) (5 s)
= 12.5 m/s
v3 = v2 + a3 t
= 12.5 m/s + (0 m/s
2
) (5 s)
= 12.5 m/s
v4 = v3 + a3 t
= 12.5 m/s + (5 m/s
2
) (5 s)
= 37.5 m/s .
006 (part 6 of 8) 10.0 points
Choose the correct plot for velocity vs time.
1.
0 5 10 15 20
−40.0
−20.0
0.0
20.0
40.0
time (s)
Velocity (m/s)
2.
0 5 10 15 20
−40.0
−20.0
0.0
20.0
40.0
time (s)
Velocity (m/s)
3.
0 5 10 15 20
−40.0
−20.0
0.0
20.0
40.0
time (s)
Velocity (m/s)
4.
0 5 10 15 20
−40.0
−20.0
0.0
20.0
40.0
time (s)
Velocity (m/s)
5.
0 5 10 15 20
−40.0
−20.0
0.0
20.0
40.0
time (s)
Velocity (m/s)
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