Top view of a block on a table is shown `(g = 10 m//s^(2))`

Find out the acceleration of the block.

In the figure shown, find out acceleration of each block.

In the figure ( g = 10 m//s^(2) ) .Acceleration of 2kg block is :

In the figure shown, F is in newton and t in seconds, Take g =10 m//s^(2) . (a) Plot acceleration of the block versus time graph. (b) Find force of friction at, t =2_(s) and t =8_(s).

In the figure shown, F is in newton and t in seconds, Take g =10 m//s^(2) . (a) Plot acceleration of the block versus time graph. (b) Find force of friction at, t =2_(s) and t =8_(s).

Block C shown in figure-1.79 is going down at acceleration 2 m//s^(2) . Find the acceleration of blocks A and B.

A100 N force acts horizontally on a block of 10 kg placed on horizontal rough table of coefficient of friction mu=0.5g=10m//s^2 . What is the acceleration of the block ?

A force F=75 N is applied on a block of mass 5 kg along the fixed smooth incline as as shown in the figure. Here gravitational acceleration g = 10 m s^(-2) . The acceleration of the block is

A force F=75 N is applied on a block of mass 5 kg along the fixed smooth incline as as shown in the figure. Here gravitational acceleration g = 10 m s^(-2) . The acceleration of the block is

A block of mass 50 kg is kept on another block of mass 1 kg as shown in figure. A horizontal force of 10 N is applied on the 50kg block. (All surface are smooth). Find (g = 10 m//s^(2)) Acceleration of block A and B.

A block of mass 2kg released from the top of a smooth fixed wedge inside the elevator which is moving downward with an acceleration a = 5 m/s^2 as shown in the figure.The value of acceleration of the block w.r.t wedge will be (g = 10 m/s^2)