Question

unfortunately, when the crate reaches a height h = 13.2 m above the ground, the crewman steps on a slick patch of ice and slips. the crate immediately accelerates toward the ground, dragging the hapless crewman across the ice and toward the edge of the cliff. assume that the ice and pulley are frictionless, and that the rope is long enough to allow the crate to hit the ground before the crewman slides over the side of the cliff. at what speed v1 will the crate hit the ground? at what speed v2 will the crewman hit the bottom of the ravine? assume no air resistance.

Explanation:

Step1: Apply conservation of mechanical - energy for the crate - hitting - ground part

The initial state has only potential energy for the crate of mass $m_1$ at height $h$, $U = m_1gh$. The final state just before hitting the ground has only kinetic energy $K=\frac{1}{2}m_1v_1^{2}$. By conservation of mechanical energy $m_1gh=\frac{1}{2}m_1v_1^{2}$.

Step2: Solve for $v_1$

Cancel out $m_1$ from both sides of the equation $m_1gh=\frac{1}{2}m_1v_1^{2}$, we get $v_1=\sqrt{2gh}$. Given $h = 13.2\ m$ and $g = 9.8\ m/s^{2}$, then $v_1=\sqrt{2\times9.8\times13.2}=\sqrt{258.72}\approx16.1\ m/s$.

Step3: For the crew - man hitting the bottom of the ravine

The crew - man has the same speed as the crate when the crate hits the ground, $v_0 = v_1$. Then the crew - man is in free - fall motion. Using the kinematic equation $v_2^{2}=v_0^{2}+2gH$, where $H$ is the height of the cliff (not given in the problem, assume the cliff height is also $h$ for simplicity, since the crew - man will fall the same height as the crate's initial height above the ground in the non - air - resistance case). Since $v_0 = v_1=\sqrt{2gh}$, then $v_2^{2}=2gh + 2gh=4gh$, and $v_2=\sqrt{4gh}=2\sqrt{gh}$. Substituting $h = 13.2\ m$ and $g = 9.8\ m/s^{2}$, we have $v_2=\sqrt{4\times9.8\times13.2}=\sqrt{517.44}\approx22.7\ m/s$.

Answer:

$v_1\approx16.1\ m/s$
$v_2\approx22.7\ m/s$