The force F = `2t^(2)` is applied on the 2 kg block. Then find out the work done by this force in 2sec. Initially at time t = 0, block is at rest.

A small block of mass 20kg rests on a bigger block of mass 30kg , which lies on a smooth horizontal plane. Initially the whole system is at rest. The coefficient of friction between the blocks is 0.5 . The horizontal force F=50N is applied on the lower block. a. Find the work done by fricitonal force on upper block and on the lower block in t=2s . b. Is the magnitude of work done by the frictional force on upper and lower blocks same? c. Is the work done by the frictional force on the upper block converted to heat or mechanical energy or both?

A small block of mass 20kg rests on a bigger block of mass 30kg , which lies on a smooth horizontal plane. Initially the whole system is at rest. The coefficient of friction between the blocks is 0.5 . The horizontal force F=50N is applied on the lower block. a. Find the work done by fricitonal force on upper block and on the lower block in t=2s . b. Is the magnitude of work done by the frictional force on upper and lower blocks same? c. Is the work done by the frictional force on the upper block converted to heat or mechanical energy or both?

A force F= 2t^2 is applied to the cart initially at rest. The speed of the cart at t = 5 s is -

Figure shows two blocks tied by a string. A variable force F =5t is applied on the block. The coefficient of friction for the blocks are 0.6 and 0.5 respectively. Find the frictional force between blocks and ground as well as tension in the string at t = 2s

A force F= 2t^2 is applied to the cart of mass 10 Kg, initially at rest. The speed of the cart at t = 5 s is -

A block of mass 2.0 kg kept at rest on an inclined plane of inclination 37^0 is pulledup the plane by applying a constant force of 20 N parallel to the incline. The force acts for one second. A. show that the work done by the applied force does not exceed 40 J. b. find the work work done by the applied force is 40 J. c. Find the kinetic energy of he block at the instant the force ceases to act. Take g=10 m/s^2 .

A force of F=0.5 N is applied on lower block as shown in figure. The work done by lower block on upper block for a displacement of 3 m of the upper block with respect to ground is (Take, g=10 ms^(-2) )

A force of F=0.5 N is applied on lower block as shown in figure. The work done by lower block on upper block for a displacement of 3 m of the upper block with respect to ground is (Take, g=10 ms^(-2) )

At t = 0 , a constant force is applied on 3 kg block. Find out maximum elongation in spring in cm.