Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. a horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
The Newton’s second law of motion is defined as,
F = m× a
⇒ a = F / m
Where
‘a’ is the acceleration of the system
‘F’ is the force applied
‘m’ is the mass of the system.
Given,
Horizontal force applied on the body, F = 600 N
Mass of the body A, mA = 10 kg
Mass of the body B, mB = 20 kg
The mass of the system, M = m �A+mB=10+20 kg= 30 kg
Thus, the acceleration ‘a’ of the system is given as,
⇒ a = 20 m/s2
(i) When the force is applied to the body A,
Using the newton’s second law of motion, The equation of motion is given as,
F – T= mA a
⇒ T= F – mA a
⇒ T= 600 N – 10 kg × 20 m/s2
⇒ T= 600 N – 200 N = 400 N
Tension when the force is applied on body A is 400 N.
(ii) When the force is applied to the body B
Using the newton’s second law of motion, the equation of motion is given as,
F – T= mB a
⇒ T= F – mB a
⇒ T= 600 N – 20 kg × 20 m/s2
⇒ T= 600 N – 400 N = 200 N
Tension when the force is applied on body B is 200 N.