Answer to Question #91356 in Mechanics | Relativity for Michelle

Question

A block of mass m starts at rest and slides down a frictionless circular ramp from a height h. At the bottom, it hits a massless spring with spring constant k and in addition begins to experience a frictional fore. The coefficient of kinetic friction is given by mu. (a) What is the speed of the block at the bottom of the ramp just before it hits the spring? (b) Find the total horizontal force on the mass after it hits the spring as a function of the coordinate x given in the diagram.

Answer 1

(a)

We can write the law of energy conservation as

$$mgh=\frac { mv^2} {2} (1)$$

We get using (1)

$$v= \sqrt {2gh}$$

(b)

The total horizontal force on the mass is equal to

$$kx-µmg$$