A particle of mass M moves along a horizontal x axis from x=0 to x=L . The coefficient of kinetic friction varies as a function of x as μ_k(x)=μ_0-α x , where μ_0 , α are constants of appropriate dimensions, so that

Question

Quantrex Academy · Accepted Answer

Correct answer: A. (1)/(2)

Given the coefficient of kinetic friction

\mu_k(x) = \mu_0 - \alpha x

At

x = L

,

\mu_k(L) = 0 \Rightarrow \mu_0 - \alpha L = 0 \Rightarrow \alpha = \dfrac{\mu_0}{L}

The frictional force acting on the particle is

f_k = \mu_k(x) Mg = \left(\mu_0 - \dfrac{\mu_0}{L}x\right)Mg

The magnitude of work done by the frictional force is given by:

|W| = \int_{0}^{L} f_k dx

|W| = \int_{0}^{L} \left(\mu_0 - \dfrac{\mu_0}{L}x\right) Mg dx

|W| = Mg \left[ \mu_0 x - \dfrac{\mu_0 x^2}{2L} \right]_{0}^{L}

|W| = Mg \left( \mu_0 L - \dfrac{\mu_0 L^2}{2L} \right) = \dfrac{1}{2} \mu_0 MgL

Comparing this with

n \mu_0 MgL

, we get

n = \dfrac{1}{2}

Answer:

\dfrac{1}{2}

A particle of mass M moves along a horizontal x axis from x=0 to...

Options

Correct answer

Step-by-step solution

More from Work, Power and Energy