Question

A block is in a lift. The lift has an acceleration a⁰ in downward direction. The work done by the normal reaction calculated by a person standing on ground is

Answer 1

The work done by the normal reaction is non-positive (zero or negative).

Answer 2

The normal force ($N$) acts upwards, while the displacement ($d$) of the block is downwards. The work done by the normal force is given by $W_N = N \cdot d \cdot \cos(180^\circ) = -N \cdot d$. The net force on the block in the downward direction is $mg - N = ma_0$. Therefore, the normal force is $N = m(g - a_0)$. For the block to remain in contact with the lift, $N \ge 0$, which implies $g \ge a_0$.

Case 1: $0 \le a_0 < g$ In this case, $N = m(g - a_0) > 0$. The work done is $W_N = -m(g - a_0)d$, which is negative (since $d > 0$).

Case 2: $a_0 \ge g$ In this case, the required normal force $N = m(g - a_0) \le 0$. Physically, the normal force cannot be negative. If $a_0 > g$, the block will lose contact, and $N=0$. If $a_0 = g$, $N=0$. In either subcase, $N=0$. Therefore, the work done by the normal force is $W_N = -0 \cdot d = 0$.

Combining both cases, the work done by the normal reaction force is always non-positive (zero or negative).