Question

a block of mass In = 2.5 kg slides head on
into a spring of spring constant k = 320 N/m. When the block
stops, it has compressed the spring by 7.5 cm. The coefficient of kinetic
friction between block and floor is 0.25. While the block is in
contact with the spring and being brought to rest, what are (a) the
work done by the spring force and (b) the increase in thermal energy
of the block-floor system? (c) What is the block's speed just
as it reaches the spring?

Answer 1

When the block reaches a stop, it has potential energy\[1/2kx ^{2}\] due to the spring. This is the work done by the spring in helping to stop the block. (a)

Friction is also consistently applying a force\[f = muR\]on the block, helping to stop it.
Since 'Work Done = Force x Distance' the work done by friction is equal to
\[W _{friction} = x.mu.R\] This is the energy dissipated to thermal energy in the system (assuming sound and internal friction in the spring are negligible) (b)

All the energy from (a) and (b) came into the system as kinetic energy of the block equal to\[1/2 mv ^{2}\]Since the block came to a complete stop,\[KE_{original} = E _{potential} + E _{thermal}\]Solve for v in the kinetic energy and you should have your original velocity.

Hope this makes sense. It's my first post! ...as a result I'm not completely sure I'm correct and am very open to correction :)