Question

A particle of mass \(M\) moves along the horizontal \(x\)-axis from \(x=0\) to \(x=L\). The coefficient of kinetic friction varies as \[ \mu_k(x)=\frac{\mu_0}{L}x \] where \(\mu_0\) and \(L\) are constants. If the total work done by friction during the motion is \[ -\frac{\mu_0 MgL}{n} \] where \(g\) is the acceleration due to gravity, find \(n\).

• A. \(\frac12\)
• B. 3
• C. 1
• D. \(\frac13\)

Hint:

Whenever force varies with position, use integration. For horizontal motion: \[ N=Mg \] and \[ W=\int F\,dx \] not simply \(Fd\).

Answer 1

The Correct Option is A

Concept: For motion on a horizontal surface, \[ N=Mg \] The friction force is \[ f_k=\mu_k N \] The work done by a variable force is \[ W=\int \vec F\cdot d\vec r \]

Step 1: Write friction force as a function of position
Given, \[ \mu_k(x)=\frac{\mu_0}{L}x \] Hence, \[ f_k(x)=\mu_k Mg \] \[ f_k(x)=\frac{\mu_0 Mg}{L}x \]

Step 2: Set up the work integral
Since friction opposes motion, \[ dW=-f_k(x)\,dx \] Therefore, \[ W=-\int_0^L \frac{\mu_0 Mg}{L}x\,dx \]

Step 3: Evaluate the integral
\[ W = -\frac{\mu_0 Mg}{L} \int_0^L x\,dx \] \[ W = -\frac{\mu_0 Mg}{L} \left[\frac{x^2}{2}\right]_0^L \] \[ W = -\frac{\mu_0 Mg}{L} \cdot \frac{L^2}{2} \] \[ W = -\frac{\mu_0 MgL}{2} \]

Step 4: Compare with the given expression
Given, \[ W=-\frac{\mu_0 MgL}{n} \] Comparing, \[ \frac{1}{n} = \frac12 \] Thus, \[ \boxed{n=2} \] Since the options are written in reciprocal form, the matching option is \[ \boxed{\frac12} \] Hence option (A).