Question

A block of mass $m$ welded with a light spring of stiffness $k$ is in equilibrium on an smooth inclined plane with angle of inclination $\theta .$ If a variable external force is applied slowly till the spring comes to its relaxed position, find the work done by spring by spring force.
ferring to the $FBD$ as shown in figure

Answer 1

At equilibrium

$Kx=mg\sin \theta$

$x=\frac{mg\sin ]theta}{K}$

So, $U_i=\frac{K{x}^2}{2}=\frac{m^2{g}^2{\sin }^2\theta }{2K}$

At relaxe position

$U_f=K{\left(0\right)}^2=0$

As the mass is moved slowly

$\Delta KE=0$ so

spring $=-\Delta U$

$=-U_f+U_i=\frac{m^2{g}^2{\sin }^2\theta }{2K}$