Question

A small block of mass m is pushed against a spring with spring constant k and held in place with a catch. The spring compresses an unknown distance x. When the catch is removed, the block leaves the spring and slides along a frictionless circular loop of radius R. When the block reaches the top of the loop, the force of the loop on the block (the normal force) is equal to twice the gravitational force on the mass. How far was the spring initially compressed?

Answer 1

Answer: x = sqrt 7mR/k

Explanation:

Answer 2

Answer:

The spring was initially compressed an distance: $x=\sqrt{\frac{7mgR}{k} }$.

Explanation:

We need to apply the conservation of energy law to find the kinetic energy at the top of the loop. First we need to find the initial state as: $E_{1}=E_{k1} +E_{U1} =\frac{kx^{2} }{2}$ where E1 is the mechanical energy at the state 1, Ek1 is the initial kinetic energy that in this case is Ek1=0 and Eu1 is the spring's potential energy and is Eu1=(1/2)kx^2. Then We need to find the final state as:$E_{2}=E_{k2}+E_{U2}=\frac{mv_{top} ^{2} }{2}+2mgR$. Remember that the surface is frictionless so $E_{1}=E_{2}$, mechanical energy is conserved. Finally we can replace and getting: $2mgR+\frac{mv_{top} ^{2} }{2}=\frac{kx^{2} }{2}$. On other side we need to apply Newton's Second Law and using the free body diagram (see attached) of the block at the top of the loop we get:$F=m*a=-mg-N=-\frac{mv_{top} ^{2} }{R}$. In this problem the force of the loop on the block is equal to twice the magnitude gravitational force on the block so N=2mg. Now solving this we can get:$3mg=\frac{mv_{top} ^{2} }{R}$ and replacing this we will find:$2mgR+\frac{3mgR}{2} =\frac{kx^{2} }{2}$ and solving for x, we can find the spring compressed distance like: $\frac{7mgR}{2} =\frac{kx^{2} }{2}$ so:$x=\sqrt{\frac{7mgR}{k} }$