Solution to Question 9
1. Analysis of Minimum Work Condition
The problem asks for the minimum work $W_{\min}$ required to drag the bar entirely out of the frictionless region into the rough region. To minimize the work done by the external agent, the process must be carried out quasi-statically, meaning the kinetic energy of the bar remains negligible ($\Delta K \approx 0$) throughout the motion.
Let $x$ be the length of the bar that has entered the rough region. The coordinate $x$ varies from $0$ to $l$.
The forces acting on the bar are:
- The spring force $F_s = k \delta$ (where $\delta$ is the extension).
- The frictional force $f_r$ acting on the portion of the bar in the rough region.
2. Force Balance at Critical Point
The critical moment occurs just as the bar is about to leave the frictionless portion entirely ($x=l$). At this point, the spring extension is maximum, and the spring force must balance the maximum static friction to maintain equilibrium.
Let the total mass of the bar be $m$. The mass of the portion in the rough region is $\frac{m}{l}x$.
The friction force as a function of $x$ is: $$f_r(x) = \mu \left( \frac{m}{l} x \right) g$$ However, we are given $W_{\min}$ directly. Let’s use the work-energy theorem. The total work done by the agent is stored as potential energy in the spring and dissipated as heat due to friction.
$$W_{\min} = \Delta U_{spring} + W_{friction}$$
3. Calculating Work Done against Friction
The work done against friction (Heat, $Q$) is the integral of the frictional force over the displacement:
$$Q = \int_0^l f_r(x) \, dx = \int_0^l \frac{\mu m g}{l} x \, dx$$
$$Q = \frac{\mu m g}{l} \left[ \frac{x^2}{2} \right]_0^l = \frac{\mu m g l}{2}$$
At the final position ($x=l$), the spring force must balance the total friction force $f_{\max} = \mu m g$. Let the final extension of the spring be $\delta$.
$$k \delta = \mu m g \implies \delta = \frac{\mu m g}{k}$$
4. Expressing Total Work
Substituting the potential energy $U = \frac{1}{2} k \delta^2$ and Heat $Q$ into the work equation:
$$W_{\min} = \frac{1}{2} k \delta^2 + Q$$
Substitute $\delta = \frac{\mu m g}{k}$ and note that $Q = \frac{\mu m g l}{2}$ implies $\mu m g = \frac{2Q}{l}$.
$$W_{\min} = \frac{1}{2} k \left( \frac{2Q}{kl} \right)^2 + Q$$
Let’s rearrange to solve for $Q$. The expression can be rewritten in terms of the maximum friction force $F_f = \mu m g$.
$$Q = \frac{F_f l}{2} \implies F_f = \frac{2Q}{l}$$
$$W_{\min} = \frac{1}{2} \frac{F_f^2}{k} + Q = \frac{1}{2k} \left( \frac{2Q}{l} \right)^2 + Q$$
$$W_{\min} = \frac{2 Q^2}{k l^2} + Q$$
5. Solving for Heat ($Q$)
This is a quadratic equation in $Q$:
$$2Q^2 + (k l^2) Q – (k l^2 W_{\min}) = 0$$
Solving for $Q$ using the quadratic formula:
$$Q = \frac{-kl^2 \pm \sqrt{(kl^2)^2 – 4(2)(-kl^2 W_{\min})}}{4}$$
$$Q = \frac{-kl^2 + \sqrt{k^2 l^4 + 8 k l^2 W_{\min}}}{4}$$
6. Numerical Calculation
Given values:
- $l = 1.0$ m
- $k = 4.0$ N/m
- $W_{\min} = 17.5$ J
$$Q = \frac{\sqrt{(4^2)(1^4) + 8(4)(17.5)} – (4)(1^2)}{4}$$
$$Q = \frac{\sqrt{16 + 560} – 4}{4} = \frac{\sqrt{576} – 4}{4}$$
$$\sqrt{576} = 24$$
$$Q = \frac{24 – 4}{4} = \frac{20}{4} = 5 \text{ J}$$