A block of mass `10 kg` is released on a fixed wedge inside a cart which is moving with constant velocity `10 ms^(1)` towards right. There is no relative motion between block and cart. Then work done by mormal reaction on block in two seconds from ground frame will be (g=10 ms^(1))

A block of mass 10 kg is released on a fixed wedge inside a cart which is moving with constant velocity 10 ms^(-1) towards right. There is no relative motion between block and cart. Then work done by normal reaction on block in two seconds from ground frame will be ( g=10 ms^(-2) )

A block of mass 10kg is released on a fixed wedge inside a cart which is moved with constant velocity 10m//s towards right. Take initial velocity of block with respect to cart zero. Then work done by normal reaction (with respect to ground) on block in two second will be :(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=5m//s^(2) Let N be the normal reaction between block and the wedge .Then (g=10m//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 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 block of mass 1 kg is at rest elative to a smooth wedge moving leftwards with constant aasseleraton a=5m//s^(2) Let N be the noemal reaction between block and the wedge .Then (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 the block and the wedge. Then : (g = 10 m//s^2)

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

A small block of mass 1 kg is kept on a rough inclined wedge of inclination 45^@ fixed in an elevator. The elevator goes up with a uniform velocity v=2 m//s and the block does not slide on the wedge. Find the work done by the force of friction on the block in 1s . (g=10 m//s^(2))