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

Step 1: Set up the normal force.
On a horizontal surface the normal force balances weight, $N = Mg$. The particle moves from $x = 0$ to $x = L$.
Step 2: Friction as a function of position.
With $\mu_k(x) = \dfrac{\mu_0}{L}x$, the kinetic friction force is
\[ f_k(x) = \mu_k N = \frac{\mu_0 Mg}{L}\,x \]
Step 3: Friction opposes motion.
Each small displacement $dx$ does work $dW = -f_k(x)\,dx$ because friction acts backward.
Step 4: Set up the integral.
\[ W = -\int_0^L \frac{\mu_0 Mg}{L}\,x\,dx \]
Step 5: Evaluate it.
\[ W = -\frac{\mu_0 Mg}{L}\left[\frac{x^2}{2}\right]_0^L = -\frac{\mu_0 Mg}{L}\cdot\frac{L^2}{2} = -\frac{\mu_0 MgL}{2} \]
Step 6: Match to find $n$.
Comparing with $W = -\dfrac{\mu_0 MgL}{n}$ gives $n = 2$. The corresponding listed option is written as $\tfrac12$.
\[ \boxed{\tfrac12} \]