**piston ramp spring**

A piston is placed on top of a spring inside a wedge-shaped block fitted with small rollers so that it can roll down an inclined plane, as shown. When the block is at rest the spring is compressed 1.6 inches from it's equilibrium position, but when the block is rolling down the plane the spring is compressed only 1.2 inches. What is the inclination angle of the plane? (Neglect friction.)

__Solution by Toby Crisford__

By Hooke's law, the force exerted by a spring is proportional to the
displacement from its equilibrium position. Calling this constant of
proportionality *k*, we know that:

*mg* = 1.6*k*,

where *m* is the mass of the piston, and* g* is acceleration due to
gravity, since the force exerted by the spring exactly balances the weight of
the piston when they are at rest. From this,

* k* = *mg*/1.6.

Once the block is rolling down the ramp, its acceleration along the ramp will be
given by

* g sin θ,*

where *θ *is the angle of inclination, since the component of the block's
weight perpendicular to the ramp will be balanced by a reaction force, and so
only the component acting along the ramp is relevant. In turn, the component of
this acceleration in the vertical direction will be

* g sin ^{2}θ.*

Because the piston accelerates with the block, the resultant vertical force on the piston must be

but

Hence,

1 - *sin ^{2}θ* = 3/4,