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Solutions of Problem set No.3: (forces, etc.)
1. In Figure 1, suppose that “Anything You Like” represents the velocity of an automobile in
units of km per hour. What is its acceleration in m s-2?
Solution:
From the graph in Fig. 1 we have:
v0  5km/hour ; v t  55 km/hour; t 0  0; t t  5sec,
The acceleration a is the rate of increase with time of the velocity:
a
v
t

t 0
dv
.
dt
In the case of the linear graph we have :
a
dv v v t  v 0 50 km/hour 50000 m/3600sec




 10000 /3600 m/sec 2  2.778 m/sec 2
dt t
t
5 sec
5 sec
2. If the mass of the automobile in the previous question is 750 kg, what force is its engine
exerting on it in order to produce such an acceleration?
Solution:
The mass of the automobile m = 750 kg. The acceleration a = 2.778 m/sec2.
Newton’s Second Law of Motion states:
F = ma = 750 kg ×2.778 m/sec2 = 2083.5N.
3. What is the “weight” of the above-mentioned automobile, in units of: (a) kg; (b) N?
Solution:
(a): The “weight” of the automobile W =750 kg.
(b): 1kg = 9.81N. The “weight” of the automobile W =(750 ×9.81) N =7357.5N
4. The planet Mars has a mass which is 10.7% that of the Earth, and a radius that is 53.2% that of
the Earth. Calculate the value of g, the acceleration due to gravity, on the surface of Mars.
Solution:
The acceleration due to gravity at Earth’s surface g = GM / R2 = 9.81 ms-2.
The mass M* of the Mars equals 0.107 M and the radius of the Mars R* equals 0.532 R. From this follows
that the acceleration due to gravity at Mars’s surface g* = GM* / (R*)2 = G x 0.107M / (0.532R)2 =
= [0.107/(0.532)2 ] × GM/R2= [0.107/(0.532)2] × g = [0.107/(0.532)2] ×9.81 ms-2  3.71ms-2.
5. If a person weighs 90 kg on Earth, how much would he weigh on Mars?
Solution:
The person weight W = 90 kg = (90 × 9.81) N = 882.9 N at Earth’s surface. This means that the person’s
mass is 90 kg and his weight at Mars’s surface W* = mg* =90 kg × 3. 71 ms-2 = 333.9N=
=(333.9/9.81)kg  34kg.