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 `m`. 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

When the plank is pulled to the right, bar also moves to the right with the plank, and elastic cord elongates. Hence, a tension is developed in it.
Since the bar starts sliding over the plank when inclination of the elastic cord becomes equal to `theta` with the vertical, therefore at this moment, friction acting between the bar and the plank is at its limiting value.

Let the force constant of the elastic cord be `K`. Then tension in it will be equal to `T=KDeltal`, where `Deltal` is elongation of the cord. It is equal to `Deltal=(l_0sectheta-l_0)`
`T=K*l_0(sectheta-1)`
Now considering FBD of the bar, for vertical equilibrium,
`N+Tcostheta=mg`
`N=(mg-Tcostheta)`
For horizontal equilibrium, `muN=T sintheta`
`mu(mg-Tcostheta)=Tsintheta`
Substituing, `T=K.l_0(sectheta-1)`
`K=(mumg)/(l_0(sectheta-1)(sintheta+mucostheta))`
Since there is not slipping between the bar and plank up to this moment, therefore no energy is lost against friction.
Since the only force responsible for rightward displacement of the bar and hence for elongation of the cord is friction, therefore whole of the work done by the friction is used to elongate the cord. In fact it is stored in the cord in the form of its deformation energy.
Hence, 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 is equal to `1/2K(Deltal)^2`.
`impliesW=1/2(mumgl_0(sectheta-1))/((sintheta+mucostheta))`