Work-energy Theorem With Friction

Question

A 2 kg block slides across a rough horizontal floor with an initial speed of 6 m/s and gradually comes to rest. If the coefficient of kinetic friction is 0.3, over what distance does the block travel before stopping?

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When friction is the only horizontal force, the work-energy theorem states that the negative work done by friction equals the loss of kinetic energy: $-\mu m g\, d = 0 - 	frac{1}{2}mv^2$. The mass cancels, leaving the stopping distance independent of mass and equal to $d = v^2/(2\mu g)$. Substituting $v = 6	ext{ m/s}$, $\mu = 0.3$, and $g = 10	ext{ m/s}^2$ gives $d = 36/(2 	imes 0.3 	imes 10) = 36/6 = 6	ext{ m}$. The option 3 m forgets the factor of two from the kinetic energy expression. The option 12 m doubles the result by misplacing that same factor. The option 9 m uses the speed linearly rather than squared. A key insight is that the stopping distance scales with the square of the initial speed, so doubling the launch speed quadruples the distance covered before the block halts. The lost kinetic energy is not destroyed but converted into thermal energy at the sliding interface, fully consistent with the broader principle of energy conservation. This is the standard NCERT application of friction as a dissipative force. As a plausibility check, the dimensions of $v^2/(\mu g)$ reduce to metres, and a block losing speed over a few metres is physically sensible.

Work-energy Theorem With Friction

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