Solution Set
11.
FN
fk
m = 50 kg
vi = 5m/s
vf = 0 m/s
Fg
Δs = 15 m
a) KEi = ½ mvi2 = ½ (50kg)(5m/s)2 = 625J
KEf = ½ mvf2 = ½ (50kg)(0)2 = 0J
b) Wnet = WN +Wg+Wk = ΔKE = KEf – KEi
Since the angle between the normal force and the displacement is 90o, WN = 0
Since the angle between gravity and the displacement is 90o, Wg = 0
Therefore Wk = -KEi = - 625J
That makes sense since the system initially had 625 J of energy in it and friction removes all of
that capacity from the system since it brings the system to rest!
c) 𝑊𝑘 = |𝑓𝑘 ||∆𝑠|𝑐𝑜𝑠𝜃𝑘
−625𝐽 = |𝑓𝑘 ||15𝑚|𝑐𝑜𝑠180
|𝑓𝑘 | = 41.7 𝑁
d) fk = µkFn
Need to first find FN
Up is positive
Fnet,y = may
FN – Fg = may
FN = Fg = mg = (50kg)(9.8 N/kg) = 490N
µk = fk/FN = 41.7N/490N = 0.085
12
FA
FN
FAy
θA
fk
m = 40 kg
vi = 0 m/s (assumed)
FA = 200N
µk = 0.20
Fg
FAx
Δs = 10 m
Some prep work first
Fg = mg = (40kg)(9.8 N/kg) = 392N
FAy = FAsinθA = 200Nsin30 = 100 N
a) Up is positive
Fnet,y = may
|𝐹𝑁 | + |𝐹𝐴𝑦 | − |𝐹𝑔 | = 𝑚𝑎𝑦
|𝐹𝑁 | + |100𝑁| − |392𝑁| = 0
|𝐹𝑁 | = 292𝑁
b) fk = µkFn =(0.2)(292N) = 58.4N
c) i) 𝑊𝐴 = |𝐹𝐴 ||∆𝑠|𝑐𝑜𝑠𝜃𝐴
𝑊𝐴 = |200𝑁||10𝑚|𝑐𝑜𝑠30 = 1732𝐽
ii) 𝑊𝑔 = |𝐹𝑔 ||∆𝑠|𝑐𝑜𝑠𝜃𝑔 = (392𝑁)(10𝑚)𝑐𝑜𝑠90 = 0𝐽
iii) 𝑊𝑁 = |𝐹𝑁 ||∆𝑠|𝑐𝑜𝑠𝜃𝑁 = (292𝑁)(10𝑚)𝑐𝑜𝑠90 = 0𝐽
iv) 𝑊𝑘 = |𝑓𝑘 ||∆𝑠|𝑐𝑜𝑠𝜃𝑘 = (58.4𝑁)(10𝑚)𝑐𝑜𝑠180 = −584𝐽
d) Wnet = WA + Wg + WN + Wk = 1732J + 0J +0J -584J = 1148J
e) Wnet = ΔKE = KEf – KEi
1148J = ½ mvf2 – 0
1148J = ½ (40kg)vf2
Vf = ± 7.58 m/s but reject the negative bc looking for speed
17)
FN
FA
FAy
FAx
θA
Δs = 15 m
Fg
m = 120 kg
FA = 200N
Some prep work first
Fg = mg = (120kg)(9.8 N/kg) = 1176N
FAy = FAsinθA = 200Nsin37 = 120.4N
a) Up is positive
Fnet,y = may
|𝐹𝑁 | + |𝐹𝐴𝑦 | − |𝐹𝑔 | = 𝑚𝑎𝑦
|𝐹𝑁 | + |120.4𝑁| − |1176𝑁| = 0
|𝐹𝑁 | = 1055.6𝑁
b) 𝑊𝑁 = |𝐹𝑁 ||∆𝑠|𝑐𝑜𝑠𝜃𝑁 = (1055.6𝑁)(15𝑚)𝑐𝑜𝑠90 = 0𝐽
Remember, you really don’t even need to do the plug in here as long as you cite that the angle is
90. With the magic of copy/paste, I don’t mind “rewriting” it but that’s not an efficient use of
your time. I just wanted to remind you how to find the normal force when there is a force acting
at an angle
c) 𝑊𝑔 = |𝐹𝑔 ||∆𝑠|𝑐𝑜𝑠𝜃𝑔 = (1176𝑁)(15𝑚)𝑐𝑜𝑠90 = 0𝐽
d) 𝑊𝐴 = |𝐹𝐴 ||∆𝑠|𝑐𝑜𝑠𝜃𝐴 = 200𝑁(15𝑚) cos 37 = 2395.9𝐽
e) Wnet = ΔKE = KEf – KEi
Wg + WN + WA = ½ mvf2 – ½ mvi2
0 + 0 + 2395.9 J = ½ (120 kg)vf2 – ½ (120 kg)(3 m/s)2
2395.9 J = 60vf2 – 540 J
vf = +/- 7.00 m/s = 7.00 m/s
I want to do some variations on the same problem on the next page
I) If there were a coefficient of friction of 0.10, what would be the final speed?
II) What would the coefficient of friction needed for the object’s speed to be 1 m/s finally?
I) Let’s not reinvent the wheel since the above problem did a lot of the work for us. All we need
to do is find the force of friction and then the work done by friction.
fk = µkFn
fk = (0.10)(1055.6 N)
fk = 105.6 N
𝑊𝑘 = |𝑓𝑘 ||∆𝑠|𝑐𝑜𝑠𝜃𝑘
𝑊𝑘 = (105.6)|15𝑚|𝑐𝑜𝑠180
𝑊𝑘 = −1584 𝐽
All we need to do is steal our work from above and add a friction work term.
Wnet = ΔKE = KEf – KEi
Wg + WN + WA + Wk = ½ mvf2 – ½ mvi2
0 + 0 + 2395.9 J + (-1584J) = ½ (120 kg)vf2 – ½ (120 kg)(3 m/s)2
811.9 J = 60vf2 – 540 J
vf = +/- 4.75 m/s = 4.75 m/s
II) Notice here that in this variation, the final speed is lower than the initial speed. That means
the net work will end up being NEGATIVE
Wnet = ΔKE = KEf – KEi
Wg + WN + WA + Wk = ½ mvf2 – ½ mvi2
0 + 0 + 2395.9 J + Wk = ½ (120 kg)(1 m/s)2 – ½ (120 kg)(3 m/s)2
2395.9 J + Wk = 60 J – 540 J
2395.9 J + Wk = -480 J
Wk = - 2875.9 J
𝑊𝑘 = |𝑓𝑘 ||∆𝑠|𝑐𝑜𝑠𝜃𝑘
−2875.9 𝐽 = |𝑓𝑘 ||15𝑚|𝑐𝑜𝑠180
|𝑓𝑘 | = 191.7 𝑁
fk = µkFn
191.7 N= µk(1055.6 N)
µk = 0.18
13. I know we did this problem together, but in order to do 14 and 15 quickly, I will recap our
discussion from that day.
FN
FA
θ
Fg
m = 200 kg
FA = 2000 N
θ = 37o
vi = 0 (implied)
Fgy
Fgx
For this problem, I really don’t care what the magnitude of the normal force is because there is
no friction in this problem. Yet, I know that problem 14 has friction so I will just find this now.
Some prep work first then
Fg = mg = (200kg)(9.8 N/kg) = 1960N
Fgy = Fg cos θ = 1960N cos 37 = 1565.3N
Off the incline is positive
Fnet,y = may
|𝐹𝑁 | + |𝐹𝑔𝑦 | = 𝑚𝑎𝑦
|𝐹𝑁 | + |1565.3𝑁| = 0
|𝐹𝑁 | = 1565.3𝑁
Remember, all of this prep work (except calculating Fg) was totally unneeded for problem 13. It
is set up for 14.
a) 𝑊𝐴 = |𝐹𝐴 ||∆𝑠|𝑐𝑜𝑠𝜃𝐴 = (2000N)(4 m)cos 0 = 8000J
b) 𝑊𝑔 = |𝐹𝑔 ||∆𝑠|𝑐𝑜𝑠𝜃𝑔 = (1960𝑁)(4𝑚) cos(90 + 37) = −4718.2𝐽
c) 𝑊𝑁 = |𝐹𝑁 ||∆𝑠|𝑐𝑜𝑠𝜃𝑁 = (𝐹𝑁 )(4𝑚)𝑐𝑜𝑠90 = 0𝐽
d) Wnet = WA + Wg + WN = 8000J -4718.2J +0 = 3281.8J
e) Wnet = ΔKE = KEf – KEi
3281.8J = ½ m vf2 – 0 = ½(200kg) vf2
Vf = ±5.73 m/s but reject the negative since speed
14. We are going to steal liberally from our work on 13
FN
FA
fk
θ
Fg
m = 200 kg
FA = 2000 N
θ = 37o
vi = 0 (implied)
Fgy
Fgx
Now there is friction with µk =0.20
Parts a – c don’t change because the work done by those other forces is not affected by adding
friction.
So we should find fk and then Wk
fk = µkFn = 0.20(1563.5N) = 312.7N
𝑊𝑘 = |𝑓𝑘 ||∆𝑠|𝑐𝑜𝑠𝜃𝑘 = (312.7𝑁)(4𝑚)𝑐𝑜𝑠180 = −1250.8𝐽
So now the net work equation changes because we add a new term
Wnet = WA + Wg + WN + Wk = 8000J -4718.2J +0 -1250.8 J= 2031J
Wnet = ΔKE = KEf – KEi
2031J = ½ m vf2 – 0 = ½(200kg) vf2
Vf = ±4.51 m/s but reject the negative since speed
It should make total sense that the speed is less!
13. So this is still largely the same problem except that FA is not directed parallel the the incline
but horizontally. This means (due to alternate interior angles) that FA makes a 37o angle with the
displacement. Now there is a stupid typo in the problem. The applied force was supposed to stay
2000N. I will do the problem both ways to compensate. First with 2000N
FN
FA
θ
Fg
m = 200 kg
FA = 2000 N
θ = 37o
vi = 0 (implied)
Fgy
Fgx
For FA = 2000N
a) 𝑊𝐴 = |𝐹𝐴 ||∆𝑠|𝑐𝑜𝑠𝜃𝐴 = (2000N)(4 m)cos 37 = 6389.1J
No changes with normal force and gravity
d) Wnet = WA + Wg + WN = 6389.1J -4718.2J +0 = 1670.9J
e) Wnet = ΔKE = KEf – KEi
1670.9J = ½ m vf2 – 0 = ½(200kg) vf2
Vf = ±4.09 m/s but reject the negative since speed
For FA = 1000N
a) 𝑊𝐴 = |𝐹𝐴 ||∆𝑠|𝑐𝑜𝑠𝜃𝐴 = (1000N)(4 m)cos 37 = 3194.6J
No changes with normal force and gravity
d) Wnet = WA + Wg + WN = 3194.6J -4718.2J +0 = -1523.6J
e) Wnet = ΔKE = KEf – KEi
-1523.6J = ½ m vf2 – 0 = ½(200kg) vf2
Vf = WAAAAAIT. I’m about to take the square root of a negative number. That means that my
final speed will be an imaginary number!!!
Did we do something wrong? Actually no, WE didn’t. The problem is with the problem. What
this means is that the object never is able to actually make it to the top of the incline. This
applied force is not big enough to accelerate this object up the incline. If we were to calculate the
x component of both the applied force and the x component of gravity, we would see the x
component of gravity would be bigger. Therefore this object would accelerate DOWN the incline
and not UP the incline as the problem assumes. So as long as the initial speed were zero, you
would not have even begun to go up the incline. Now, if the initial speed were say 5 m/s going
up the incline to begin with….
e) Wnet = ΔKE = KEf – KEi
-1523.6J = ½ m vf2 – ½ m vi2 = ½(200kg) vf2 – ½ (200kg)(5 m/s)2
-1523.6 = 100 vf2 – 2500J
976.4 = 100 vf2
Vf = ±3.12 m/s and reject the negative bc speed
So, the lesson here is that if you get the square root of a negative number, DON’T FUDGE IT.
That imaginary number informs you about something PHYSICAL…that this situation only could
happen in your imagination ;)
6. P = 2000 W
m = 400 kg
hf = 10 m (taking the ground as h = 0)
𝑃=
𝑊
𝑡
Here, the work being done is a work done against gravity so
W = ∆PEg
W = (400 kg)(9.8 N/kg)(10 m)
W = 39,200 J
2000 𝑊 =
39,200 𝐽
𝑡
t = 19.6 s
8.
A) 1 kWhr = (1000 W)(3600 s) = 3,600,000 J
B)
P = 1200 W = 1.2 kW
You could convert everything to seconds but that isn’t useful here.
Work = (Power)(time)
Work = (1.2 kW)(24 hrs)
Work = 28.8 kWhr
Price = (Energy supplied)(Cost)
Price = (28.8 kWhr)(14.7 cents/kWhr)
Price = 423.36 cents = $4.23