A horizontal plane supports a plank with a bar of mass `m` placed on it and attached by a light elastic non-deformed cord of length `l_0`, to a point O. The coefficient of friciton between the bar and the plank is equal to `mu`. The plank is slowly shifted to the right until the bar starts sliding over it. It occurs at the moment when the cord derivates from the vertical by an angle `theta`.

Find the work that has been performed by the moment by the frictional force acting on the bar in the reference frame fixed to the plane.

Explanation

`N+T cos theta=mg` ...(i)
`F_(f)=T sin theta=0` ... (ii)
`:.F_(f)=T sin theta` ...(iii)
and `N=mg-T cos theta` ...(iv)
Further, `T=K(l-l_(0))=kl_(0)(sec theta -1)`
`x=l_(0)tan theta`
`dx=l_(0)sec^(2)theta d theta`
Work done, `dW=vec(F)_(f).dvec(s)=T sin theta (dx)`,
`W= int T sin theta(dx)`
`W=k l_(0)^(2)int_(theta=0)^(30^(@))(tan theta-sin theta)sec^(2)theta d theta= kl_(0)^(2)[((tan theta)^(2))/( 2)-(1)/(cos theta)]_(0)^(30^(@))=kl_(0)^(2)xx(0.012)`
when `theta=30^(@), kl_(0)(sec 30^(@)-1)=(0.2 mg)/(1/2+0.2xx(sqrt(3)/(2))`
`rArr kl_(0)=1.92mg`
`:.W=(0.023)mgl_(0)=0.092 J`.