Question

A student uses a compressed spring of force constant 22 N/m to shoot a 0.0075 kg eraser across a desk. The magnitude of the force of friction on the eraser is 0.042 N. How far along the horizontal desk will the eraser slide if the spring is initially compressed 0.035m

Answer 1

The initial elastic potential energy stored in the spring is:
$U= \frac{1}{2}k x^2 = \frac{1}{2}(22 N/m)(0.035 m)^2=0.0135 J$

Then, the spring is released, all this potential energy is converted into kinetic energy of the eraser:
$U= K=0.0135 J$

Then, the force of friction does a work to stop the eraser in this motion, and this work is equal to
$W=Fd$
where F is the magnitude of the friction force and d is the  distance covered by the eraser.

For energy conservation, the work  done by the friction force must be equal to the initial energy of the eraser, so:
$K=W=Fd$
and so we find d:
$d= \frac{K}{F}= \frac{0.0135 J}{0.042 N}=0.32 m$