Under the action of force, 2kg body moves such that its position x as a function of time t is given by x= (t^(3))/(30x is in metre and t in second. Calculate the work done by the force in the first 2 second

Under the action of force, 2kg body moves such that its position x as

1.	Under the action of force, 2kg body moves such that its position x as a function of time t is given by x= (t^(3))/(30x is in metre and t in second. Calculate the work done by the force in the first 2 second
Answer» Solution :From work energy theorem `W= Delta KE` `x= t^(3)//3 therefore` velocity `v= (dx)/(dt) = t^(2)` At t=0, `v_(i) = 0^(2)= 0` At t= 2, `v_(F)= 2^(2)= 4m//s` word DONE `W= (1)/(2) m(v_(f)^(2)-v_(i)^(2))` `=(1)/(2) xx 2(4^(2)-0)= 16J`