Solution
Given:
- Mass of stone, $m = 5 \text{ kg}$
- Linear mass density of chain, $\lambda = 0.5 \text{ kg/m}$
- Initial velocity, $v_0 = 20\sqrt{6} \text{ m/s}$
- Acceleration due to gravity, $g = 10 \text{ m/s}^2$
Part (a): Length of chain $L = 60 \text{ m}$
Assume the stone rises to a height $h$. Since the total length $L=60 \text{ m}$ is quite large, let us test the assumption that the chain is partially on the ground ($h < L$).
Applying Conservation of Mechanical Energy:
$$ K_i + U_i = K_f + U_f $$
At the highest point, velocity is zero ($K_f = 0$).
The potential energy at height $h$ consists of:
- PE of the stone: $mgh$
- PE of the suspended chain portion (mass $\lambda h$): The center of mass of the vertical part is at $h/2$. Thus, $U_{\text{chain}} = (\lambda h) g (h/2) = \frac{\lambda g h^2}{2}$.
Equation:
$$ \frac{1}{2}mv_0^2 = mgh + \frac{\lambda g h^2}{2} $$
Rearranging for $h$:
$$ \lambda g h^2 + 2mgh – mv_0^2 = 0 $$
Substituting values: $\lambda = 0.5, m = 5, v_0^2 = (20\sqrt{6})^2 = 2400$
$$ (0.5)(10)h^2 + 2(5)(10)h – (5)(2400) = 0 $$
$$ 5h^2 + 100h – 12000 = 0 $$
Dividing by 5:
$$ h^2 + 20h – 2400 = 0 $$
Solving the quadratic equation:
$$ h = \frac{-20 + \sqrt{400 – 4(1)(-2400)}}{2} = \frac{-20 + \sqrt{400 + 9600}}{2} $$
$$ h = \frac{-20 + \sqrt{10000}}{2} = \frac{-20 + 100}{2} = 40 \text{ m} $$
Since $h = 40 \text{ m} < 60 \text{ m}$, our assumption was correct.
Answer: 40 m
Part (b): Length of chain $L = 30 \text{ m}$
Since the previous result ($40 \text{ m}$) is greater than $L=30 \text{ m}$, the chain will be completely lifted off the ground.
Energy Equation:
$$ \frac{1}{2}mv_0^2 = U_{\text{stone}} + U_{\text{chain}} $$
At height $h$, the stone is at $h$. The chain of length $L$ hangs vertically from the stone.
Center of mass of the chain is at height $h – L/2$.
Mass of chain $M_c = \lambda L$.
$$ \frac{1}{2}mv_0^2 = mgh + (\lambda L)g\left(h – \frac{L}{2}\right) $$
$$ \frac{mv_0^2}{2g} = mh + \lambda L h – \frac{\lambda L^2}{2} $$
$$ h(m + \lambda L) = \frac{mv_0^2}{2g} + \frac{\lambda L^2}{2} = \frac{mv_0^2 + \lambda g L^2}{2g} $$
$$ h = \frac{mv_0^2 + \lambda g L^2}{2g(m + \lambda L)} $$
Substituting values ($L=30, v_0^2=2400$):
$$ h = \frac{5(2400) + (0.5)(10)(30^2)}{2(10)(5 + 0.5 \times 30)} $$
$$ h = \frac{12000 + 5(900)}{20(5 + 15)} = \frac{12000 + 4500}{20(20)} $$
$$ h = \frac{16500}{400} = \frac{165}{4} = 41.25 \text{ m} $$
Answer: 41.25 m