University of Hail, Physics Department
Physics 112: Problems For Chapters 7 and 8
KE 
W  F  d  Fd cos  .
1
mv 2 .
2
U  W .
U = mgy=mgh.
A  B  Ax Bx  Ay By  Az .Bz
Emec =KE+ U.
Wnc =  Emec.
K = W. Fs=-kx
Us 
1 2
kx
2
Chapter 7
1) Calculate the kinetic energy of an automobile of mass m= 80×104g traveling at 54.0 km/h.
KE 
1
mv 2
2
2) A force
where m= 800 kg and
v
54
 15 m / s
3.6
so KE = 0.5×800×152 = 9×104 J
F   5.0 Ni  3.0 N j  10 N k is applied to a particle of mass 4.5 kg. Calculate the work
done by F as the particle is moved trough displacement d   300 i  250 j  cm .
W  F  d  Fx d x  Fy d y  Fz d z  (5)  3.0  3  (2.5)  10  0   22.5J
3) The block, of mass m = 8.0 kg was initially at rest, is pulled, by the force F = 50 N, to the
right for a distance d= 2.5 m as shown in figure.
a) Calculate the work done by F.
F
W  F  d  Fd cos 
W = 50×2.5cos30
= 108.3 J
m
 
d
b) Calculate the final speed v of the block.
K = W ,
Kf – Ki = W
4v2 = 108.3, this gives : v2 =27.1 and v  27.1  5.2 m / s
1
mv 2  0  108.3
2
4) A 10 kg block is dragged 20 m by a parallel force of 26 N. If k = 0.20 calculate the net work
done on the block.
The net work: Wnet = WF + Wfk
⃗⃗⃗⃗ = 𝐹𝑑 cos 𝜃 = 𝐹𝑑𝑐𝑜𝑠0
WF = 𝐹⃗ ∙ 𝑑
= 26 × 20 × (1) = 520 𝐽
Wfk = 𝑓⃗𝑘 ∙ ⃗⃗⃗⃗
𝑑 = 𝑓𝑘 𝑑 cos 𝜃 = 𝑓𝑘 𝑑𝑐𝑜𝑠180
fk
10 Kg
F=26N
20 m
k
fk =kFN, where FN is the normal force. In this case FN = mg =10×9.8 = 98 N and the frictional
force fk =0.20×98 = 19.6 N
Wfk = = 19.6 × 20 × (−1) = −392 𝐽 and the net work is: Wnet = 520 – 392 =128 J
5)
A 7 kg block is dragged 15 m by a force F=20 N that make an angel of 35o with the
20 N
35o
15 m
horizontal. If k = 0.20.
7 kg
a) Calculate the work done by the force F.
k
WF = 𝐹⃗ ∙ ⃗⃗⃗⃗
𝑑 = 𝐹𝑑 cos 𝜃 = 𝐹𝑑𝑐𝑜𝑠35 = 20 × 15 × 0.82 = 246 𝐽.
b) Calculate the work done by the frictional force.
fk =kFN where FN is the normal force. We will calculate the normal force first.
y
From the free body diagram “FBD”
on the y-axis: FN + Fsin – W = 0
FN
this gives : FN = W - Fsin
F
= mg – 20sin35 = 7×9.8 – 11.5 = 57 N.
 
m
So: fk =kFN = 0.2×57= 11.4 N
⃗⃗⃗⃗ = 𝑓𝑘 𝑑 cos 𝜃 = 𝑓𝑘 𝑑𝑐𝑜𝑠180 = 11.4 × 15 × (−1) = −171 𝐽
Wfk = 𝑓⃗𝑘 ∙ 𝑑
W
c) Calculate the net work done on the block.
The net work is: Wnet = WF + Wfk = 246-171= 75 J
6) A spring stretches 20 cm when a 980 g mass is hung from it. Determine the spring constant.
The spring force, Fs = kx, balance the weight W = mg:
Fs = W
Fs
20 cm
kx = mg
this gives
k=
𝑚𝑔
𝑥
=
0.980×9.8
0.20
= 48 𝑁/𝑚
980 g
W
x
Chapter 8
1) A 100 kg box is lifted up above the ground until its potential energy is 4.9 kJ relative to the
ground. Calculate its height above the ground.
The potential energy is: U = mgh.
4.9×103 = 100×9.8×h. This gives ℎ =
4900
980
= 5.0 𝑚
2) A stone of mass 50 kg is lifted to a height of 20 m above the ground and then released to fall
y
freely. Calculate the speed of the stone when it has fallen to the ground.
y=20 m
Conservation of mechanical energy:
Ei = Ef
KEi +Ui = KEf+ Uf
1
0 + mgy = 𝑚𝑣 2 + 0
2
This gives: v2 = 2gy and 𝑣 = √2𝑔ℎ = √2 × 9.8 × 20 = 19.8 𝑚/𝑠
y=0m
3) A ball of mass 1.5 kg is thrown down, with an initial speed v0=4.0 m/s, from the roof of a 40
m tall building, calculate its initial mechanical energy.
The mechanical energy: Emec = KE + U
1
= 2 𝑚𝑣 2 + 𝑚𝑔ℎ
= 0.5×1.5×42+1.5×9.8×40
= 600 J
vo
40 m
4) A car of mass m=400.0 kg moves with a speed of 72 km/h on the top of a hill of height 5.0 m
above the ground. Determine the mechanical energy of the car.
1
72 km/h
2
Emec = KE + U = 2 𝑚𝑣 + 𝑚𝑔ℎ.
72
We have: 𝑣 = 3.6 = 20 𝑚/𝑠 then
Emec = 0.5×400×202 + 400×9.8×5
= 99600 J
5m
5) What is the potential energy of a spring compressed by a distance of -1.2 cm? (k=100N/m)
U==
1
2
𝑘𝑥 2 = 0.5×100×(-0.012)2=0.0072 J