Solution to Question 12
1. Work Done in Trial 1 (Upright Box)
The box stands upright. The spring pushes the stack of $n$ plates upwards against the top lid. The normal force exerted by the spring balances the weight of the plates and provides the normal reaction at the lid.
Let $N_{lid}$ be the normal force at the top lid when $i$ plates remain. Let $F_s$ be the spring force.
$$N_{lid} + (i \cdot mg) = F_s \implies N_{lid} = F_s – i mg$$
The friction acts on both the top surface (against lid) and bottom surface (against next plate). When pulling the top plate ($i$-th plate), the normal force on top is $N_{lid}$ and on bottom is $N_{bot} = N_{lid} + mg$.
Total friction on $i$-th plate:
$$f_i = \mu N_{lid} + \mu N_{bot} = \mu (2 N_{lid} + mg) = \mu [ 2(F_s – img) + mg ]$$
$$f_i = \mu [ 2F_s – (2i – 1)mg ]$$
2. Work Done in Trial 2 (Inverted Box)
The box is inverted. The spring is now at the top, pushing the stack down against the “lid” (now at the bottom). Gravity now acts in the same direction as the spring force.
The normal force at the bottom lid $N’_{lid}$ must support both the spring force and the weight of the plates.
$$N’_{lid} = F_s + i mg$$
When pulling the bottom plate ($i$-th plate), the friction acts on the bottom (against lid) and top (against next plate). $N’_{top} = N’_{lid} – mg$.
Total friction on $i$-th plate:
$$f’_i = \mu N’_{lid} + \mu N’_{top} = \mu (2 N’_{lid} – mg) = \mu [ 2(F_s + img) – mg ]$$
$$f’_i = \mu [ 2F_s + (2i – 1)mg ]$$
3. Difference in Work Done ($\Delta W$)
The work done is the sum of friction forces times the displacement $l$ (edge length) for all $n$ plates. We seek the difference $\Delta W = W_2 – W_1$.
$$\Delta W = \sum_{i=1}^n (f’_i – f_i) l$$
Substituting the expressions for friction:
$$f’_i – f_i = \mu [ 2F_s + (2i-1)mg ] – \mu [ 2F_s – (2i-1)mg ]$$
$$f’_i – f_i = 2\mu mg (2i – 1)$$
Summing over $i$ from 1 to $n$:
$$\Delta W = 2\mu mg l \sum_{i=1}^n (2i – 1)$$
The sum of the first $n$ odd numbers is $n^2$.
$$\Delta W = 2 \mu m g l n^2$$
4. Calculation
We are given $\Delta W = 10$ J, $n = 20$, $l = 5$ cm $= 0.05$ m, $\mu = 0.5$, $g = 10$.
$$10 = 2 (0.5) (m) (10) (0.05) (20^2)$$
$$10 = 1 \cdot m \cdot 0.5 \cdot 400$$
$$10 = 200 m$$
$$m = \frac{10}{200} = 0.05 \text{ kg} = 50 \text{ g}$$