Question

an inclined plane of angle 20, has a spring of force constant k=500 N/m, fastened securely at the bottom so that the spring is parallel to the surface,a block of mass m =2.50 kg is placed on the plane at a distance d=0.3m from the spring. From this position ,the block is projected downwards towards the spring with speed v=0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest ?

Answer 1

By conservation of energy:

$$\dfrac 12kx^2=\dfrac 12mv^2+mgL\sin\theta,\\\space\\ x=\sqrt{\dfrac mk(v^2+2gL\sin\theta)}=0.11\text{ m}.$$