Understanding the Work-Energy Theorem
The work-energy theorem states that the net work done on an object equals its change in kinetic energy: $W_{net} = \Delta K = K_f - K_i$
In the context of vertical motion near Earth's surface, the only work done on the rock (neglecting air resistance) is by the gravitational force. This work is conservative, so we can also view this as the Conservation of Mechanical Energy: $E_i = E_f$.
Solving the Problem
Given Data:
- Weight ($W$) = 20 N. Since $W = mg$, mass $m = W/g = 20/9.8 \approx 2.04$ kg.
- Initial height $h_i = 0$ m.
- Final height $h_f = 15$ m.
- Final velocity $v_f = 25$ m/s.
- Acceleration $g = 9.8$ m/s$^2$.
Part (i): Initial Speed ($v_i$)
Using the principle of Conservation of Mechanical Energy: $K_i + U_i = K_f + U_f$ $\frac{1}{2}mv_i^2 + 0 = \frac{1}{2}mv_f^2 + mgh_f$
Divide by $m$: $\frac{1}{2}v_i^2 = \frac{1}{2}(25)^2 + (9.8)(15)$ $v_i^2 = 625 + 294 = 919$ $v_i = \sqrt{919} \approx 30.3$ m/s
Part (ii): Maximum Height ($h_{max}$)
At maximum height, the vertical velocity $v_{top} = 0$. $E_{ground} = E_{top}$ $\frac{1}{2}mv_i^2 = mgh_{max}$ $h_{max} = \frac{v_i^2}{2g} = \frac{919}{2 \times 9.8} \approx 46.89$ m