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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 mg22μ0


From the FBD:
Applying equilibrium condition along y axis,
mgcosθ=N ...(i)
Along x axis,
mgsinθf=ma ...(ii)
Given μ=μ0x
Then, f=μN=μ0xmgcosθ ...(iii)

From Eq (ii) & (iii):
mgsinθμ0xmgcosθ=ma
a=gsinθμ0xgcosθ ...(iv)
We know that a=vdvdx
Substituting in Eq (iv),
vdvdx=gsinθμ0xgcosθ
Grouping terms and integrating on both sides
vf=0vi=0vdv=x0(gsinθμ0xgcosθ)dx
(limits of velocity are both zero, since block starts from rest and finally comes to rest)
[v22]00=gsinθ[x]x0μ0gcosθ[x22]x0
0=gsinθ xμ0gcosθ 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=(01μ0) ]
ΔH=μ0mgcosθ1μ00xdx=μ0mgcos45[x22]1μ00
ΔH=μ0mg×12×12μ20
ΔH=mg22μ0

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