(b) A `10` kg weight is raised to a height of `2.0` m with an acceleration of `1.5 m//s^(2)`. Compute the work done, if `g=10 m//s^(2)`.

Calculate the work done in lifting a 300N weight to a height of 10m with an acceleration 0.5ms^(-2) . Take g= 10ms^(-2) .

A man is standing on a weighing machine placed in a lift. When stationary his weight is recorded as 40 kg . If the lift is accelerated upwards with an acceleration of 2m//s^(2) , then the weight recorded in the machine will be (g=10m//s^(2))

A coin is released inside a lift at a height of 2m from the floor of the lift. The height of the lift is 10 m. The lift is moving with an acceleration of 9 m//s^(2) downwards. The time after which the coin will strike with the lift is : (g=10 m//s^(2))

A block of mass 1 kg is at rest relative to a smooth wedge moving leftwards with constant acceleration a = 5 m//s^(-2) .Let N be the normal reation between the block and the wedge . Then (g = 10 m//s^(-2))

A man of weight 80 kg is standing in an elevator which is moving with an acceleration of 6m//s^(2) in upward direction. The apparent weight of the man will be (g=10m//s^(2))

A 2 kg block A is attached to one end of a light string that passes over an an ideal pulley and a 1 kg sleeve B slides down the other part of the string with an acceleration of 5m//s^(2) with respect to the string. Find the acceleration of the block, acceleration of sleeve and tension in the steing. [g=10m//s^(2)]

A block of mass 1 kg is at rest relative to a smooth wedge moving leftwards with constant acceleration a=5m//s^(2) Let N be the normal reaction between block and the wedge .Then (g=10m//s^(2) )

A body of mass 10 kg moving at a height of 2 m, with uniform speed of 2 m/s. Its total energy is

If acceleration of A is 2m//s^(2) to left and acceleration of B is 1m//s^(2) to left, then acceleration of C is-

A block of mass 10 kg is released on rough incline plane. Block start descending with acceleration 2m//s^(2) . Kinetic friction force acting on block is (take g=10m//s^(2) )