Rock Mass Wave Distance 57Eefe

1. **Problem 28:** Find the mass of a rock falling vertically with air resistance force $12$ N and acceleration $2.0\ \text{m/s}^2$ downwards, given gravitational field strength $g=10\ \text{N/kg}$. 2. **Formula and concepts:** The net force $F_{net}$ on the rock is given by Newton's second law: $$F_{net} = m a$$ The forces acting are weight $W = mg$ downwards and air resistance $F_{air} = 12$ N upwards. So, $$F_{net} = W - F_{air} = mg - 12$$ Since acceleration $a=2.0\ \text{m/s}^2$ downwards, $$m a = mg - 12$$ 3. **Calculate mass $m$: $$m a = mg - 12$$ $$m a - mg = -12$$ $$m(a - g) = -12$$ $$m = \frac{-12}{a - g} = \frac{-12}{2.0 - 10} = \frac{-12}{-8} = 1.5$$ 4. **Answer for problem 28:** The mass of the rock is $1.5\ \text{kg}$ (Option C). --- 5. **Problem 29:** A transverse wave with amplitude $4.0$ cm and frequency $10$ Hz travels at speed $2.4\ \text{m/s}$. Find the total distance travelled by a particle in the rope in $20$ s. 6. **Formula and concepts:** The particle moves up and down in simple harmonic motion. The distance travelled by a particle in one complete oscillation is the circumference of the motion path, which is $4$ times the amplitude (up, down, back, and return). 7. **Calculate total distance travelled:** - Frequency $f = 10$ Hz means $10$ oscillations per second. - Number of oscillations in $20$ s is $$N = f \times t = 10 \times 20 = 200$$ - Distance travelled in one oscillation is $$d_{osc} = 4 \times \text{amplitude} = 4 \times 4.0\ \text{cm} = 16.0\ \text{cm} = 0.16\ \text{m}$$ - Total distance travelled $$d_{total} = N \times d_{osc} = 200 \times 0.16 = 32.0\ \text{m}$$ 8. **Answer for problem 29:** The total distance travelled by a particle in $20$ s is $32.0$ m.