Question : Two bodies of masses 10kg and 20kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F=600N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
Doubt by Krish
Solution :
m1=10 kg
m2=20 kg
F = 600 N
i) When force F=600N is applied to A
Let a be the common acceleration of the system of two blocks.
F=(m1+m2)a
600=(10+20)a
600=30a
600/30=a
20 =a
a=20 m/s²
Now,
For body B, we can write down the equation
T1=m2a
T1=20×20
T1=400 N
ii) When force F=600N is applied to B
Let again a be the common acceleration of the system of two blocks.
F=(m1+m2)a
600=(10+20)a
600=30a
600/30=a
20 =a
a=20 m/s²
Now,
For body A, we can write down the equation
T2=m1a
T2=10×20
T2=200 N
Hence, the tension of the string in case i) is 400 N and in case ii) it is 200 N.