Answer to Question #281586 in Classical Mechanics for Nobo

Question

As shown in the figure, a very small block of mass m is pushed against a spring of spring constant

k. After some finite compression of length l, the block is released and it leaves the spring and

slides through a horizontal surface. The kinetic friction force between this block and the horizontal

surface is μk = bx, where b is a positive constant. It then slides along a frictionless circular loop

of radius R. Gravitational force is acting vertically downward and the magnitude of gravitational

acceleration is g. When the block reaches at 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.

(a) Find the work done on the block by the kinetic friction force while it travels from x = 0

to x = d along the horizontal surface.

(b) Find the tangential velocity of the block at the top point of the circular loop.

(c) Find an expression of l in terms of some or all of the following: g, d, b, k, R and m

Answer 1

(a) The work done by friction along distance d:

$$W=\mu_k mgd.$$

(b) The velocity of the block in the upper point of the loop:

$$v=\sqrt{2gR}.$$

(c) To reach that height at this velocity, the velocity of the block before entering the loop is

$$\frac12 mu^2=\frac12 mv^2+mg·2R,\\\space\\ u^2=v^2+2gR.$$

Then add this to the work done against friction:

$$\frac12kl^2=W+\frac12mu^2,\\\space\\ \frac12kl^2=\mu_kmgd+\frac12m(2gR+2gR)^2,\\\space\\ l=\sqrt{\frac mk\Big[2\mu_kgd+(4gR)^2\Big]}.$$