Introduction to the Work-Energy Theorem
The work-energy theorem states that the net work done on an object equals its change in kinetic energy ($W_{net} = \Delta KE$). In the context of gravity, the only force acting on a projectile (neglecting air resistance) is gravity, which is a conservative force. Therefore, we can express this as the conservation of mechanical energy: $KE_i + PE_i = KE_f + PE_f$.
Given Data
- Weight ($W$): 20 N
- Mass ($m$): Since $W = mg$, then $m = W/g = 20/9.8 \approx 2.04\ \text{kg}$
- Height at observed point ($h_1$): 15 m
- Velocity at observed point ($v_1$): 25 m/s
- Initial height ($h_0$): 0 m
(i) Finding the Speed at Launch ($v_0$)
Using the principle of conservation of energy between the ground ($h=0$) and the point at 15 m:
$$KE_0 + PE_0 = KE_1 + PE_1$$ $$\frac{1}{2}mv_0^2 + 0 = \frac{1}{2}mv_1^2 + mgh_1$$
Divide by mass ($m$): $$\frac{1}{2}v_0^2 = \frac{1}{2}(25)^2 + (9.8)(15)$$ $$\frac{1}{2}v_0^2 = 312.5 + 147$$ $$\frac{1}{2}v_0^2 = 459.5$$ $$v_0^2 = 919$$ $$v_0 = \sqrt{919} \approx 30.32\ \text{m/s}$$
(ii) Finding the Maximum Height ($H_{max}$)
At maximum height, the vertical velocity becomes zero ($v_{max} = 0$). We can set the initial state (at ground) equal to the final state (at peak):
$$KE_0 + PE_0 = KE_{peak} + PE_{peak}$$ $$\frac{1}{2}mv_0^2 + 0 = 0 + mgH_{max}$$
Again, cancel $m$ and solve for $H_{max}$: $$\frac{1}{2}v_0^2 = gH_{max}$$ $$H_{max} = \frac{v_0^2}{2g}$$ Using the known value $v_0^2 = 919$: $$H_{max} = \frac{919}{2 \times 9.8}$$ $$H_{max} = \frac{919}{19.6} \approx 46.89\ \text{m}$$
Summary of Intuition
- The Work-Energy Theorem allows us to bypass complex kinematic equations by looking at the energy states of the system.
- Because gravity is a conservative force, the energy added as kinetic energy at the bottom is gradually converted into potential energy as the rock rises.
- At the peak, all initial kinetic energy has been successfully converted into gravitational potential energy, which is why the velocity hits zero.