Q: A spherical steel ball released at the top of a long column of glycerin of length l falls through a distance 1/2 with accelerated motion and the remaining distance 1/2 with uniform velocity. Let t1 and t2 denote the times taken to cover the first and second half and W1 and W2 are the work done against gravity in the two halves, then compare times and work done.
Sol: Average velocity in first half of the distance < v, while in the second half the average velocity is v. Therefore, t1 > t2. The work done against gravity in both halves is mg l/2
∴t1 > t2 ∴ W1 = W2