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

Step-by-step solution

Here, the position of the object is x = 3 t 2 + 5 , therefore the velocity of the object will be, v = d x d t = d 3 t 2 + 5 d t v = 6 t + 0 From the work-energy theorem, all the work done by the force acting on it will be equal to the change in its kinetic energy, i.e., W = K E t = 5   s - K E t = 0   s W = 1 2 × 2 × 6 × 5 2 - 1 2 × 2 × 6 × 0 2 W = 900   J .