Energy bookkeeping in three stops: the top of the ramp, the bottom, and the spring at full squeeze. Friction is the one thing that takes energy out, as in Friction and Energy. Each problem here is one short scene and a chain of questions. Answer them in order: each later question uses what you found before, so a slip early on carries through. Keep full precision between steps and round only at the end.
Key idea
On a ramp the normal force is $mg\cos\theta$, so friction does $|W_f| = \mu mg\cos\theta\cdot L$. Then $$\tfrac{1}{2}mv^2 = mgL\sin\theta - |W_f|$$ at the bottom, and on a smooth floor $\tfrac{1}{2}kx^2 = \tfrac{1}{2}mv^2$.
Worked example: A crate into a spring
A $2.0\ \text{kg}$ crate slides $3.0\ \text{m}$ down a $30^\circ$ ramp ($\mu = 0.2$), then into a $400\ \text{N/m}$ spring on a smooth floor. Find (1) the work done by friction, (2) the speed at the bottom and (3) the squeeze.
- (1) $|W_f| = (0.2)(2.0)(9.8)\cos 30^\circ(3.0) = 10.2\ \text{J}$.
- (2) $\tfrac{1}{2}mv^2 = (2.0)(9.8)(3.0)\sin 30^\circ - 10.2 = 19.2\ \text{J}$, so $v = 4.38\ \text{m/s}$.
- (3) $x = \sqrt{2(19.2)/400} = 0.310\ \text{m}$.
Answer: $10.2\ \text{J}$, $4.38\ \text{m/s}$, $0.310\ \text{m}$
Common mistake
Using $\mu mg$ for friction on a slope. The ramp only presses back with $mg\cos\theta$. Check: on a vertical wall ($\theta = 90^\circ$) there would be no normal force at all.