Physics 9A HW 3 — Due 4/24 (FRIDAY), by 5pm. NOTE THE CHANGE IN DEADLINE!
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From the text. . .
Reading — Chapters 5 and 6
Some Suggested Practice Problems (not to be turned in) — 5.63, 5.64, 5.87, 5.88, 5.90, 5.91, 5.95, 5.96,
5.99, 5.100, 5.102, 5.105, 5.107, 5.109, 5.110, 5.111, 5.112, 5.113, 5.115, 5.117, 5.118, 5.120, 5.121, 5.122,
5.124, 5.126
In addition to the Mastering Physics HW3 assignment, complete these problems.
Problem 1:
A pickup truck accelerates uniformly from rest to 95km/h in 15.0s. A 25.0kg box (starting in the front
of the truck bed) accelerates at only half that rate until it hits the back of the 2.00m-long truck bed.
(a) From the perspective of a person standing stationary by the side of the road, what is the work
done by friction over the interval from when the truck begins moving to when the box hits the
back of the truck bed?
Solution:
Let’s define positive as the direction of the truck’s travel and set the origin at where the back of
the truck is at the start of the acceleration. We can calculate the acceleration of the truck and
the box:
atruck =
95km/h
26.4m/s
∆v
=
=
= 1.8m/s2
∆t
15.0s
15.0s
→
abox = 0.88m/s2
The only force on the box that does work is the friction, which is parallel to the displacement.
It is also equal to the net force, and therefore
Ff = Fnet = mbox abox = (25.0kg)(0.88m/s2 ) = 22N
Now we need the displacement of the box from when the truck begins accelerating to when it
hits the back in a reference frame attached to the ground. Thus we need the positions of both
the box and the back of the truck as functions of time.
1
xtruck (t) = atruck t2
2
1
xbox (t) = abox t2 + x0,box
2
→
1
1
atruck t2 = abox t2 + x0,box
2
2
where x0,box = 2.00m. Thus, using atruck = 2abox ,
s
r
2x0,box
2(2.00m)
1
1
t=
=
→
∆xbox = abox t2 = (0.88m/s2 )(2.1s)2 = 1.9m
2 = 2.1s
abox
0.88m/s
2
2
Thus the work is
W = F|| ∆x = (22N)(1.9m) = 42J
Note: There is an algebraic solution that will give the displacement as exactly 2 meters (rather
1.9 from
rounding for
sig
figs), however
this 19:12:14
is ONLY
because of the ratio atruck = 2abox .
This study sourcethan
was downloaded
by 100000795812085
from
CourseHero.com
on 04-24-2022
GMT -05:00
In general, this is the method that would need to be used.
https://www.coursehero.com/file/13268819/HW3-solution/
(b) The truck continues at 95km/h for some time, and then decelerates to a stop, and the box remains
in the back of the truck bed. From the perspective of a car traveling past at a constant 95km/h in
the same direction that the truck was moving, what was the work done on the box in the stopping
interval? What force did this work?
Solution:
The work done on the box in this interval changed its kinetic energy. In the given frame of
reference (using the same positive direction as in (a) ) the box begins with zero velocity and
ends with a velocity of -95km/h. This means that the work is
1
1
W = ∆KE = mbox ∆(v 2 ) = (25.0kg) (−26.4m/s)2 − (0m/s)2 = 8.7kJ
2
2
(note that I waited until the end to round to 2 sig figs). The friction force is again the force
doing the work - without friction the box would slide to the front of the truck.
(c) From the perspective of a car traveling past at a constant 95km/h in the opposite direction that
the truck was moving, what was the work done on the box in the stopping interval?
Solution:
Similarly to (b), but now the box begins with a velocity of -190km/h and ends with a velocity
of 95km/h. This means that the work is
1
1
W = ∆KE = mbox ∆(v 2 ) = (25.0kg) (26.4m/s)2 − (−52.8m/s)2 = −26kJ
2
2
Problem 2:
A horizontally oriented spring is attached to a block of mass 5.00kg sitting on a rough surface with
µk = 0.500. The spring (with a spring constant of 272N/m) is initially compressed 55.0cm. The block
is then released.
(a) Quantitatively prove that the block will pass through the equilibrium point at least once.
Solution:
The force on the block in the interval from the initial point (x = −0.550m), where the spring is
compressed and the speed is zero, to the equilibrium point for the spring (which we will denote
as x = 0), is:
F = −kx − µmg
The spring force is positive (recall that x is negative in this interval) and the friction force is
negative. The work done on the block is therefore
Z 0
1
1
W =
−kx − µmg dx = − kx2 − µmgx|0xi = kx2i + µmgxi
2
2
xi
Recall that xi is negative. If this work is positive, the block will have increased its speed, and
therefore will be going through the spring’s equilibrium point. If it is zero, or negative, this will
not be the case (instead the block will stop at or before then, respectively). We can calculate
this:
1
W = (272N/m)(−0.550m)2 + (0.500)(5.00kg)(9.81m/s2 )(−0.550m) = 27.7J
2
It’s positive, so the block passes the equilibrium of the spring.
(b) How far from the spring’s equilibrium point will the block first stop?
This study sourceSolution:
was downloaded by 100000795812085 from CourseHero.com on 04-24-2022 19:12:14 GMT -05:00
https://www.coursehero.com/file/13268819/HW3-solution/
The block will stop when its speed returns to zero. Over this interval, the friction acts in one
direction (towards the block’s starting point) and the work is zero (starting and ending at zero
kinetic energy). Thus
Z xf
1
x
−kx − µmg dx = − kx2 − µmgx|xfi = 0
W =
2
xi
1 2
1 2
2µmg
2µmg
2
2
kx + µmgxi = kxf + µmgxf
→
0 = xf +
xf − xi +
xi
2 i
2
k
k
s
µmg 2 µmg
2µmg
2
xf = −
±
xi
+ xi +
k
k
k
We choose the + root because we know that the block passes through equilibrium, and so xf
must be positive. We can now calculate xf
µmg
(0.500)(5.00kg)(9.81m/s2 )
=
= 0.0902m
→
k
272N/m
p
xf = −(0.0902m) + (0.0902m)2 + (−0.550m)2 + 2(0.0902m)(−0.550m) = 0.370m
(c) What is the block’s speed as it passes through the equilibrium point for the second time.
Solution:
We approach this as we did for part (a), though now we calculate the speed. Friction has changed
directions (opposing the sliding back towards the equilibrium point), so the work is
Z xf
1
x
−kx + µmg dx = − kx2 + µmgx|xfi
W =
2
xi
Now xi = 0.370m and xf = 0.
1
W = kx2i − µmgxi
2
We can set this equal to the change in kinetic energy
1
1
∆KE = m∆(v 2 ) = mvf2 = W
→
2
2
s
r
k 2
272N/m
vf =
xi − 2µgxi =
(0.370m)2 − 2(0.500)(9.81m/s2 )(0.370m) = 1.95m/s
m
5.00kg
Since this is a speed, it must be positive.
This study source was downloaded by 100000795812085 from CourseHero.com on 04-24-2022 19:12:14 GMT -05:00
https://www.coursehero.com/file/13268819/HW3-solution/
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