Question

A uniform solid cylinder of mass m and radius r is released from the top of a fixed rough inclined plane of inclination theta. The coefficient of friction between the cylinder and the inclined plane is mu =1/4 tan theta. Find:1. The time taken by the cylinder to reach the bottom.2. Total kinetic energy of the cylinder at the bottom.3. Total work done by friction.

Answer 1

See it is not a case of pure rolling . as value of coefficient of friction is less than required.

Answer 2

i). 3.8 $\sqrt{1 Sin}$θ = The time taken by the cylinder to reach the bottom

ii).  7.5 ml Sinθ = Total kinetic energy of the cylinder at the bottom

iii). $-\frac{1}{4} mgl$ Sinθ = Total work done by friction

Explanation:

i). Taking u = 0 and by using s = ut + $\frac{1}{2} at^{2}$

l = 0 + 1/2 * 7.5 sin θ * $t^{2}$

$t^{2} = \frac{2l}{7.5 sin∅}$

$t^{2} = \sqrt{\frac{1}{3.75 sin θ} }$

Now,

$v^{2} - u^{2} = 2as$

$v^{2}$− 0 = 2 * 7.5sinθ * l

v = 3.8$\sqrt{1 Sin}$θ

ii). Total kinetic energy of the cylinder at the bottom = $\frac{1}{2}mv^{2} }$

= 1/2 * m * 2 * 7.5 sinθ * 1

= 7.5 ml sinθ

iii) 0  = N − mgcosθ

The work done by the friction force is then;

W = $f^{1}_{0} F_{x}$dx = −μ * N * l

= -(1/4) tanθ * mgcosθ * l

= -(1/4) mglSinθ

Learn more: work done by friction

brainly.in/question/5193086