Numerical Problems
3.1 A force of 20 N moves a body with an acceleration of 2 ms – 2 . What is its mass?
(10kg)
Solution: Force = F = 20 N
Acceleration = a = 2 ms – 2
Mass = m = ?
F = ma
Or m = F / a = 20 / 2
m = = 10 kg
3.2 The weight of a body is 147 N. What is its mass?
(14.7 kg)
Solution: Weight = w = 147 N
Acceleration due to gravity = g = 10 ms – 2
Mass = m = ?
w = mg
or m = w / g
m = 147 / 10
m = 14. 7 kg
3.3 How much force is needed to prevent a body of mass 10 kg from falling?
(100 N)
Solution: Mass = m = 50 kg
Acceleration = a = g = 10 ms – 2
Force = F = ?
F = m a
F = 10 x 10
F = 100 N
3.4 Find the acceleration produced by a force of 100 N in a mass of 50 kg.
Solution: Force = F = 100 N
Mass = m = 50 kg
Acceleration = a = ?
F = m a
Or a = F / m
a = 100 / 50
a = 2 ms – 2
(24 N)
Solution: Weight = w = 20 N
Acceleration = a = 2 ms – 2
Vertically upward force (Tension) = T = ?
Fnet = T – w
Or ma = T – mg
Or ma + mg = T
Or T = m (a + g) ……………………(i)
Now, m =
m = = 2 kg
Putting the value of m in Eq.(i), we get
T = 2 (2 + 10)
= 2(12)
T = 24 N
3.6 Two masses 52 kg and 48 kg are attached to the ends of a string that passes over a frictionless pulley. Find the tension in the string and acceleration in the bodies when both the masses are moving vertically.
Solution: m1 = 52 kg and m2 = 48 kg
- Tension T = ?
- Acceleration a = ?
3.7 Two masses 26 kg and 24 kg are attached to the ends of a string which passes over a frictionless pulley. 26 kg is lying over a smooth horizontal table. 24 N mass is moving vertically downward. Find the tension in the string and the acceleration in the bodies.
(125 N, 4.8 ms – 2)
Solution: : m1 = 24 kg and m2 = 26 kg
- Tension T = ?
- Acceleration a = ?
3.8 How much time is required to change 22 Ns momentum by a force of 20 N?
(1.1s)
Solution: Change in momentum = Pf – PI = 22 Ns
Force = F = 20 N
Time = t = ?
3.9 How much is the force of friction between a wooden block of mass 5 kg and the horizontal marble floor? The coefficient of friction between wood and the marble is 0.6.
(30 N)
Solution: Mass = m = 5 kg
Coefficient of friction = µ = 0.6
Force of friction = FS = ?
FS = µ R (where R = mg)
FS = µ mg
FS = 0.6 x 5 x 10 = 30 N
3.10 How much centripetal force is needed to make a body of mass 0.5 kg to move in a circle of radius 50 cm with a speed 3 ms – 1?
(9 N)
Solution: Mass = m = 0.5 kg
Radius of the circle = r = 50 cm = = 0.5 m
Speed = v = 3 ms – 1
Centripetal force = Fc = ?







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