Question

Two bodies A and B of masses 10kg 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 do (i)A (ii) B along the direction of the string. What is the tension in the string in each case?

Answer 1

Horizontal force, F = 600 N

Mass of body A, m1 = 10 kg

Mass of body B, m2 = 20 kg

Total mass of the system, m = m1 + m2 = 30 kg

Using Newton’s second law of motion, the acceleration (a) produced in the system can be calculated as:

F = ma

therefore a=f/m=600/20m/s2
When force F is applied on body A:
the answer is in image



The equation of motion can be written as:

F – T = m1a

∴T = F – m1a

= 600 – 10 × 20 = 400 N … (i)

When force F is applied on body B:

The equation of motion can be written as:

F – T = m2a

T = F – m2a

∴T = 600 – 20 × 20 = 200 N … (ii)

I hope this will help you

Answer 2

Hii friend,

# Answer-
i) T1 = 400 N
ii) T2 = 200 N

# Explaination-
# Given-
m1 = 10 kg
m2 = 20 kg
F = 600 N

# Solution-
Here, common acceleration is given by
a = F/(m1+m2)
a = 600/(10+20)
a = 20 m/s

(i) When force is applied to body A-
T = F - m1a
T = 600 - 10×20
T = 600 - 200
T = 400 N

(ii) When force is applied to body B-
T = F - m2a
T = 600 - 2p×20
T = 600 - 400
T = 200 N

Hope that helped you...