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NEET 2026 (Re-neet) · Work by a Variable Force

Work by a Variable Force question from NEET 2026 (Re-neet)

Q1NEET 2026 · RE-NEETMedium

Work by a Variable Force

A particle of mass MM moves along a horizontal xx axis from x=0x = 0 to x=Lx = L. The coefficient of kinetic friction varies as a function of xx as μk(x)=μ0−αx\mu_k(x) = \mu_0 - \alpha x, where μ0,α\mu_0, \alpha are constants of appropriate dimensions, so that μk(L)=0\mu_k(L) = 0. The total work done by the frictional force during the motion is nμ0MgLn\mu_0 MgL, where gg is the acceleration due to gravity. The value of nn is:

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Answer: (a)

  1. μk(L)=0⇒α=μ0L\mu_k(L) = 0 \Rightarrow \alpha = \frac{\mu_0}{L}
  2. Wf=−∫0LμkMg dx=−Mg[μ0L−αL22]=−Mg[μ0L−μ0L2]W_f = -\int_0^L \mu_k Mg\,dx = -Mg\left[\mu_0 L - \frac{\alpha L^2}{2}\right] = -Mg\left[\mu_0 L - \frac{\mu_0 L}{2}\right]
  3. Wf=−12μ0MgLW_f = -\frac{1}{2}\mu_0 MgL, magnitude 12μ0MgL\frac{1}{2}\mu_0 MgL
  4. So the answer is (a) n=12n = \frac{1}{2}.
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