Question

# A small mass starts sliding down a fixed inclined plane of inclination 45∘ with the horizontal. The coefficient of friction is μ=μ0x where x is the distance through which the mass slides down and μ0 is a constant. Find the heat produced during half the distance travelled by the particle before it stops.

A
mg2μ0
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B
mg22μ0
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C
mgμ0
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D
mg2μ0
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Solution

## The correct option is B mg2√2μ0 From the FBD: Applying equilibrium condition along y− axis, mgcosθ=N ...(i) Along x− axis, mgsinθ−f=ma ...(ii) Given μ=μ0x Then, f=μN=μ0xmgcosθ ...(iii) From Eq (ii) & (iii): ⇒mgsinθ−μ0xmgcosθ=ma ∴a=gsinθ−μ0xgcosθ ...(iv) We know that a=vdvdx Substituting in Eq (iv), vdvdx=gsinθ−μ0xgcosθ Grouping terms and integrating on both sides ∫vf=0vi=0vdv=∫x0(gsinθ−μ0xgcosθ)dx (limits of velocity are both zero, since block starts from rest and finally comes to rest) ⇒[v22]00=gsinθ[x]x0−μ0gcosθ[x22]x0 ⇒0=gsinθ x−μ0gcosθ x22 ⇒x=2sinθμ0cosθ=2μ0tanθ Putting θ=45∘, x=2μ0tan45∘=2μ0 is the distance travelled along the inclined plane before the mass stops. Magnitude of work done by friction in small displacement (dx) is dW=−fdx =−μmgcosθ dx=−(μ0x)mgcosθ dx Heat produced=Work done by friction ∴ΔH=∫1μ00(μ0xmgcosθ) dx [For half distance, limits are x=(0→1μ0) ] ΔH=μ0mgcosθ∫1μ00xdx=μ0mgcos45∘[x22]1μ00 ΔH=μ0mg×1√2×12μ20 ∴ΔH=mg2√2μ0

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