Two blocks of masses `m_1` and `m_2` inter connected by a spring of stiffness `k` are placed on a horizontal surface. If a constant horizontal force F acts on the block `m_1` it slides through a distance x whereas `m_2` remains stationary. If the coefficient of friction between all contacting surfaces is `mu`, find the speed of the block `m_1` as the function `x`.

Explanation

Since the block `m_2` does not move force acting on `m_2` do not perform work. As block do not move in vertical direction, hence work done by normal reaction and weight will be zero. Then, the force `F_(sp)`, F and `F_(k)` do work on the block

Let v=speed of `m_1`
Applying `W-E` theorem,
`W_(sp)+W_N+W_(gr)+W_(f_k)+W_(ext)=DeltaK`
`(-1/2kx^2)+(0)+(0)-muNx+Fx=(m_1v^2)/(2)`.
where `N=m_1g`
`v=sqrt(2(F/m_1-mug)-k/m_1x^2)`