In one dimensional motin, a 1 kg body experiences a force, which is linear function of time t viz. `F=2t` acting in the direction of motion. What is the work done by the force in the first 4 second?

A force acts on a 2 kg object so that its position is given as a function of time as x=3t^(2)+5. What is the work done by this force in first 5 seconds ?

A force acts on a 2 kg object so that its position is given as a function of time as x=3t^(2)+5. What is the work done by this force in first 5 seconds ?

Under the action of force, 2 kg body moves such that its position x as a function of time t is given by x=(t^(3))/(3) , x is in metre and t in second. Calculate the work done by the force in the first 2 second.

A body of mass 10 kg moves according to the relation x = t^(2) + 2t^(3) . The work done by the force in the first 2s is.

Under the action of a force, a 2 kg body moves such that its position x as a function of time is given by x =(t^(3))/(3) where x is in meter and t in second. The work done by the force in the first two seconds is .

Under the action of a force, a 2 kg body moves such that its position x as a function of time is given by x =(t^(3))/(3) where x is in meter and t in second. The work done by the force in the first two seconds is .