Question

An initially stationary 3.8 kg object accelerates horizontally and uniformly to a speed of 14 m/s in 3.6 s. In that 3.6 s interval, how much work is done on the object by the force accelerating it?

What is the instantaneous power due to that force at the end of the interval?

What is the instantaneous power due to that force at the end of the first half of the interval?

Answer 1

The acceleration of the object is given by substracting the velocities and dividing by the time interval, and that gives us a = 3,88 m/s². Therefore, the force accelerating the object is just F = m.a = 14,77 N. Now, to compute the work done by this force we need to know the distance the object moved in those 3,6s. To do that we use kinematic equation X_f = X_o + vt + at²/2, and we get form there Xf = 28m. So the work done by the force is just W = F.d = 413,56 J.

The instantaneous power is given by P = F.v, so at the end of the interval the power is P = 206,78W.

To compute the instantaneous power at the end of the first half of the interval, we need to know the speed of the object in half of the interval (that is in t = 1,8s), we calculate that from the equation v_f = v_o + at = 6,98 m/s. Therefore, the power is P = F.v = 103,15 W.