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Homework 7 Solution!
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A 1 kilogram block is pushed with a force of 5.0 newtons and let go after 2.0 meters. The first 1
meter of surface is frictionless. The second 1 meter of surface is a rough patch and has a
coefficient of kinetic friction of 0.60. The third 1 meter of surface is frictionless. The fourth 1 meter
of surface is exactly like the second 1 meter. The surface ends with a frictionless 30° ramp.!
stop
push
let go
30°
5N
1m
1m
1m
µk=0.6
1m
µk=0.6
!
a.! How far up the ramp does the block travel before stopping?!
Over the first rough patch,!
b.! How much total work is done on the block?!
c.! How much time does the block spend over it?!
d.! What are the block’s initial and final velocities?!
e.! What is the change in the kinetic energy of the block?!
Over the second rough patch,!
f.! How much total work is done on the block?!
g.! How much time does the block spend over it?!
h.! What are the block’s initial and final velocities?!
i.! What is the change in the kinetic energy of the block?!
Practice Problems Standard Difficulty!
Chapter 7 Energy!
2, 4, 8, 9, 10,11, 18, 19, 24, 25, 26, 28, 30, 61, 62, 65, 68!
Chapter 7 Power!
43, 46, 47, 48, 50, 71, 74, 75, 77, 80!
Chapter 7 Spring!
33, 34, 66!
Practice Problems Challenging Difficulty!
Chapter 7!
83, 84, 85!
!
page 1
a.! I will apply the work-energy theorem from the beginning to the end.!
Wtotal = ΔK !
You can’t draw one free body diagram because the forces come on and go off over time. You
have to identify all of the work that is done on the block. Work is done by the push force, by
friction on both patches, and by gravity. The normal force does zero work.!
The work done by the push force is!
Wpush = Fpush Δr(+1) = +(5)(2) = +10 J !
The work done by the friction over the first patch is!
Wfriction,1 = Ffriction Δr(−1) = −µkmgΔr = −(0.6)(1)(9.8)(1) = −5.88 J !
The work done by the friction over the second patch is the same!
Wfriction,2 = −5.88 J !
The work done by gravity is!
Wgravity = mgΔr cos120° = (1)(9.8)x(−0.5) = −4.9x !
The change in kinetic energy is zero since the block starts at rest and ends at rest.!
Together,!
10 − 5.88 − 5.88 − 4.9x = 0 !
−4.9x = 1.76
⇒ x = −0.359 m !
This means the block never get to the ramp so the distance up the ramp is zero.!
b.! The total work done on the block over the first patch is the sum of the work done by the push
force and friction.!
Wpush = Fpush Δr(+1) = +(5)(1) = +5 J !
Wfriction,1 = −5.88 J !
The total work done on the block over the first patch is –0.88 J.!
c.! While the block is over the patch, the net force is the push force minus the friction force which
is a constant. Therefore, the acceleration is constant and we can apply the constant
acceleration equations of motion to this. Here is what we know.!
a=?
ti = 0
tf = ?
xi
xf
vi
vf
We need more information. The net force tells us the acceleration.!
ΣF = Fpush − Ffriction = 5 − (0.6)(1)(9.8) = −0.88 N = (1)a !
a = −0.88
m
s2
page 2
We can also find the initial speed over the first patch by looking at the what happened before
reaching the patch. The block was pushed from rest with 5 N of force over 1 meter. The workenergy theorem for this part looks like this.!
1
⇒ (5)(1) = (1)v f2
2
Wtotal = ΔK
⇒ vf =
10 m/s !
Here is what we have now.!
a = –0.88
ti = 0
tf = ?
xi
xf
vi
vf
I will now use the third equation of motion to find the final velocity.!
Δ(v 2 ) = 2aΔx
2
⇒ v f2 − 10 = 2(−0.88)(1) ⇒ v f = 2.8705 m/s !
And use the first equation of motion to find the final time.!
Δv = aΔt
⇒ 2.8705 − 10 = (−0.88)t f
⇒ t f = 0.33152 s !
Here are the final values.!
a = –0.88
ti = 0
tf = 0.332
xi
vi
xf
vf
Here is a check using the second equations of motion.!
1
Δx = vi Δt + aΔt 2
2
1
⇒ 1 = ( 10)t f + (−0.88)t f2
2
⇒
1
(0.88)t f2 − ( 10)t f + 1 = 0 !
2
t f = 0.33152 s !
The block spends 0.332 second over the first patch.!
d.! The block’s initial velocity is 3.16 m/s and its final velocity is 2.87 m/s over the first patch.!
e.! The total work done on the block is –0.88 J. Using the work-energy theorem, the change in
the kinetic energy is also –0.88 J.!
We can check this using the velocities above.!
1
ΔK = (1)(2.87052 − 10) = −0.88 J !
2
f.! Now, since the block never goes through the second patch, we need to know how far it actually
goes if we want the displacement over the second patch. After the first patch, the remaining
kinetic energy in the block is!
10 − 5.88 = 4.12 J !
Therefore, over the second patch, friction will do this much work to stop the block. The total work
done over the second path is just –4.12 J.
page 3
g.! Again, the net force is constant and so is the acceleration. Here is what we know.!
a=?
ti = 0
tf = ?
xi
xf
vi
vf
We need more information. The net force tells us the acceleration.!
ΣF = −Ffriction = −(0.6)(1)(9.8) = −5.88 N = (1)a !
a = −5.88
m
s2
!
This is all we need.!
a = –5.88
ti = 0
tf = ?
xi
xf
vi
vf
I will use the first equation of motion to find the final time.!
Δv = aΔt
⇒
− 2.8705 = (−5.88)t f
⇒ t f = 0.48818 s !
The block stops over the second patch after 0.488 second.!
h.! The initial velocity of the block over the second patch is 2.87 m/s and its final velocity is zero.!
i.! Again, the change in the kinetic energy is just the total work done on it which is –4.12 J.!
We can check this using the velocities above.!
1
ΔK = (1)(0 − 2.87052 ) = −4.12 J !
2
page 4