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A hollow sphere of mass 100g is released from the top of an inclined plane of inclination $ϕ (= 45^{0})$ and height h ( = 8 m). The coefficient of friction between the plane and the sphere is 0.2. The work done (in J) by the force of friction as the sphere rolls down the inclined plane is $w_{1}$ in rotation about centre of mass and $w_{2}$ in translation of centre of mass. Find the value of $w_{1} - w_{2} (g = 10 m / s^{2})$

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answer is 2.2.

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Detailed Solution

$μ_{\min}$ for pure rolling $= \tan θ (\frac{β}{1 + β}) = 0.4$
Given $μ = 0.2 < 0.4$
$W_{1} = \frac{12 μ^{2} m g}{1 - μ} = 0.6$

$W_{2} = - 8 μ m g = - 1.6$

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