A body of mass 0.5 kg travels in a straight line with velocity `v=a x(3/2)` where `a=5 m(1/2)s^-1`. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

Certain force acting on a 20 kg mass changes its velocity from 5ms ^-1 to 2ms^-1 . Calculate the work done by the force.

A force F acting on the object varies with distance x as shown in fig.The force is in N and x is in m.The work done by the force in moving the object from x =0 to x=6 m is : .

A body of mass 2 kg travels according to the law x (t) = pt + qt^2 + rt^3 where p = 3 ms^-1 q = 4 ms^-2 and r = 5 ms^-3 . the force acting on the body at t = 2 seconds is

The displacement x of a body of mass 10 kg moving in one dimension under the action of a constant force is related to time by the equation t = sqrtx + 3 . Find the work done by the force in first 5 seconds.

A constant force acts on object of mass 5 kg for a duration of 2s It increases the object's velocity from 3m/s to 7m/s Find the magnitude of the applied force. Now if the force were applied for a duration of 5s what would be the final velocity of object?

What is the distance of the point (-1,2,1) from the plane 2x-3y+4z+5=0.