Question

he block is released, and it slides 2.0 m (from the point at which it is released) across a horizontal surface before friction stops it. What is the coefficient of kinetic friction between the block and the surface?

Answer 1

Answer:

0.245

Explanation:

When the block is released, the initial elastic potential energy stored in the spring is entirely converted into kinetic energy of the block.

Therefore, we can calculate the initial speed of the block:

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

where the term on the left is the potential energy and where the term on the right is the kinetic energy, and where

k = 4500 N/m is the spring constant

x = 8.0 cm = 0.08 m is the compression of the spring

m = 3.0 kg is the mass of the block

v is the initial velocity

Solving for v,

$v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(4500)(0.08)^2}{3.0}}=3.1 m/s$

Then, after the block is released, all its kinetic energy is converted into thermal energy as the block slows down, due to friction. Therefore, the work done by friction is equal to the initial kinetic energy of the block.

The force of friction is

$F=\mu mg$

where

$\mu$ is the coefficient of friction

$g=9.8 m/s^2$ is the acceleration of gravity

So the work done by it is (in magnitude)

$W=Fd=\mu mg d$

where

d = 2.0 m is the distance covered

Therefore,

$\frac{1}{2}mv^2 = \mu mg d$

And solving for $\mu$,

$\mu = \frac{v^2}{2gd}=\frac{3.1^2}{2(9.8)(2.0)}=0.245$