Block sliding on incline compressing spring

Given: $m = 10$ kg, $\theta = 30^\circ$, $\mu = 0.20$, $d = 4$ m, $k = 2000$ N/m.

   \
    \  block
     \[#]
      \ \  4 m
       \ \_____
        \  /\/\/\ spring
         \/_______

Forces along the incline

• Component of weight down the slope: $mg\sin30^\circ = 10 \times 9.81 \times 0.5 = 49.05$ N
• Normal reaction: $N = mg\cos30^\circ = 84.96$ N
• Friction (up the slope): $\mu N = 0.20 \times 84.96 = 16.99$ N
• Net driving force: $49.05 - 16.99 = 32.06$ N

Speed just before touching the spring

Work-energy over 4 m: $0 + F_{net}\, d = \tfrac12 m v^2$

$$v = \sqrt{\frac{2 \times 32.06 \times 4}{10}} = 5.06\ \text{m/s}$$

Maximum compression x

The block moves $(4 + x)$ along the incline from rest to rest (momentarily stopped at maximum compression). The spring does negative work $-\tfrac12 kx^2$.

$$F_{net}(4 + x) - \tfrac12 k x^2 = 0$$ $$1000x^2 - 32.06x - 128.23 = 0$$ $$x = \frac{32.06 + \sqrt{32.06^2 + 4(1000)(128.23)}}{2000} = 0.374\ \text{m}$$

(The negative root is rejected.)

Answer: Maximum compression = 374 mm (0.374 m); speed before touching the spring = 5.06 m/s.