Question bank/NEET UG/Physics

NEET UGPhysicsModerateRe-NEET 2026

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 μk(x)=μ0−αxμ_k(x) = μ_0 - αx, where μ0μ_0, α are constants of appropriate dimensions, so that μk(L)=0μ_k(L) = 0. The total work done by the frictional force during the motion is nμ0MgLnμ_0MgL, where g is the acceleration due to gravity. The value of n is:

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Sol. Given μk(x)=μ0−αxμ_k(x) = μ_0 - αx and μk(L)=0=μ0−αL⇒μ0=αLμ_k(L) = 0 = μ_0 - αL \Rightarrow μ_0 = αL
Magnitude of work done by frictional force, Wf=∫dWfW_f = \int dW_f
where dWf=fkdx=μkNdx=μkMgdx=(μ0−αx)MgdxdW_f = f_k dx = μ_k N dx = μ_k Mg dx = (μ_0 - αx)Mg dx
∴Wf=∫0Lμ0Mgdx−∫0LαMgxdx=μ0Mg[x]0L−αMg[x22]0L∴ W_f = \int_{0}^{L} μ_0 Mg dx - \int_{0}^{L} αMgx dx = μ_0Mg[x]_{0}^{L} - αMg\left[\frac{x^2}{2}\right]_{0}^{L}
⇒nμ0MgL=μ0MgL−(μ0L)Mg(L22)=μ0MgL2\Rightarrow nμ_0MgL = μ_0MgL - \left(\frac{μ_0}{L}\right)Mg\left(\frac{L^2}{2}\right) = \frac{μ_0MgL}{2}
⇒n=12\Rightarrow n = \frac{1}{2}

Source: geekyrakshit/indian-exam-questions / reneet2026-code50

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