Source: Principles of Physics
Problem
A block of cheese of mass $m$ lies on the floor of an elevator cab of mass $M$. The cab is pulled upward by a cable.
- Through a distance $d_1$, the normal force on the block from the floor has a constant magnitude $F_{N1}$. How much work $W_{c1}$ is done on the cab by the force from the cable?
- Through a distance $d_2$, the work done on the cab by the constant force from the cable is $W_{c2}$. What is the magnitude of the normal force $F_{N2}$ on the block?
[Q1] $W_{c1} = \frac{(M+m)F_{N1}d_1}{m}$ [Q2] $F_{N2} = \frac{m W_{c2}}{d_2(M+m)}$
Let $a$ be the upward acceleration of the elevator system, and $T$ be the tension in the cable. We apply Newton's second law. The positive direction is upward.
[Q1] First, find the acceleration $a_1$ by analyzing the forces on the cheese.
$$F_{N1} - mg = ma_1 \implies a_1 = \frac{F_{N1} - mg}{m}$$Next, find the cable tension $T_1$ by analyzing the forces on the combined system (cab + cheese).
$$T_1 - (M+m)g = (M+m)a_1$$Substitute the expression for $a_1$:
$$T_1 = (M+m)(g+a_1) = (M+m)\left(g + \frac{F_{N1} - mg}{m}\right) = (M+m)\left(\frac{mg + F_{N1} - mg}{m}\right) = \frac{(M+m)F_{N1}}{m}$$The work done by the cable is the tension multiplied by the distance.
$$W_{c1} = T_1 d_1 = \frac{(M+m)F_{N1}d_1}{m}$$[Q2] First, find the cable tension $T_2$ from the work done $W_{c2}$ over the distance $d_2$.
$$W_{c2} = T_2 d_2 \implies T_2 = \frac{W_{c2}}{d_2}$$Next, find the acceleration $a_2$ by analyzing the forces on the combined system.
$$T_2 - (M+m)g = (M+m)a_2 \implies a_2 = \frac{T_2}{M+m} - g = \frac{W_{c2}}{d_2(M+m)} - g$$Finally, find the normal force $F_{N2}$ by analyzing the forces on the cheese.
$$F_{N2} - mg = ma_2$$ $$F_{N2} = m(g+a_2) = m\left(g + \frac{W_{c2}}{d_2(M+m)} - g\right) = \frac{m W_{c2}}{d_2(M+m)}$$