In the system of pulleys shown what should be the value of `m_(1)` such that 100 gm remains at rest : (Take `g = 10 m//s^(2)`)

In the system shown in the adjoining figure, the acceleration of the 1 kg mass is m//s^(2) . (take g = 10 m//s^(2) )

The coefficient of static friction between the block of 2 kg and the table shown in figurwe is mu_s=0.2 What should be the maximum value of m so tht the blocks do not move? Take g=10 m/s^2. The string and the pulley are light and smooth.

The coefficient of static friction between the block of 2 kg and the table shown in figurwe is mu_s=0.2 What should be the maximum value of m so tht the blocks do not move? Take g=10 m/s^2. The string and the pulley are light and smooth.

The coefficient of static friction between the block of 2 kg and the table shown in figurwe is mu_s=02. What should be the maximum value of m so tht the blocks do not move? Take g=10 m/s^2. The string and the pulley are light and smooth.

For the equilibrium of the system shown, the value of mass m should be

For the equilibrium of the system shown, the value of mass m should be

In the arrangement shown in figure, what should be the mass of block A so that the system remains at rest. Also find force exerted by string on the pulley Q. (g = 10 m//s^(2))

The system starts from rest and A attains a velocity of 5 m/s after it has moved 5 m towards right, Assuming the arrangement to be frictionless every where and pulley & string to be light, the value of the force F applied on A is (take g = 10 ms^(-2) )

If pulley and all strings as shown are massless, then tension T_(2) = 100 N . Calculate weight of the block ( g = 10 m//s^(2))