Question

Two blocks are attached with spring are placed on a horizontal smooth surface. Block a is placed in left side whose mass is 1 kilogram. Block b is in right side and its mass is 2 kilogram. A force of 5 newton is applied on 1 kilogram in left direction where as the force of 20 newton is applied on 2 kilogram in right direction. Find the maximum elongation in the spring. Spring constant is 100 n/m?

Answer 1

Suppose force F is applied on mass m1 (=2kg) which is in contact with the mass m2 (=3kg).If the normal reaction ‘R’applied by mass ‘m1′ on mass’ m2′ and vice versa. Then for mass ‘my we have

F—R=m1 a

R=m2 a

F=(m1+m2)a

a=F/(m1+m2)=100/(2+3)

=20m/s^2

Reaction between the two blocks, R=m2 a

=(3×20) N

=60 N