a uniform solid cylinder of mass m and radius r is released from the top of a rough inclined plane - Physics - System Of Particles And Rotational Motion

Question

a uniform solid cylinder of mass m and radius r is released from
the top of a rough inclined plane of inclination theta and length
of slant = l the coefficient of friction is muu=(1/4)tan theta find
time taken by cyillender to reach the bottom , total kinetic energy
of the cylinder at the bottom , total work done by friction
pls reply fast

Jyoti Pant · Accepted Answer

As the cylinder is sliding down the inclined plane, force of
friction F acts up the plane If a is acceleration produced in the
block, then net force on the cylinder down the plane is,
f=ma=mgsinθ-F=mgsinθ-μR=mgsinθ-μ mg cosθ=mg (sinθ-μ cosθ)or a=g (sinθ-14×tanθ× cosθ)=g (sinθ-14×sinθ)=34g sinθ=7
5 sinθNow,u=0 Using, s=ut+12at2l=0+12×7
5 sinθ×t2t2=2l7 5 sinθ or t=l3
75 sinθAgain,v2-u2=2asv2-0=2×7
5 sinθ×lor v=3
8l sinθKinetic energy=12mv2=12×m×2×7
5 sinθ×l=7 5 ml sinθ
Now again,N-mgcosθ=0The work done by the friction force is thenW=∫0lFxdx=-μ×N×l=-14tanθ×mgcosθ×l=-14mgl sinθ