Question

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. figure not to scale at what speed ( v_1 ) will the crate hit the ground? ( v_1 = 5.23 \text{ m/s} ) correct answer at what speed ( v_2 ) will the crewman hit the bottom of the ravine? assume no air resistance. ( v_2 = 15.33 \text{ m/s} ) incorrect answer

Explanation:

Step1: Analyze the motion of the crate

Use conservation of mechanical - energy for the crate - crewman system. The initial potential energy of the crate is converted into the kinetic energy of the crate and crewman when the crate hits the ground. Let the mass of the crate be $m_1$ and the mass of the crewman be $m_2$. The initial potential energy of the crate is $U = m_1gh$, and the final kinetic energy of the system is $K=\frac{1}{2}(m_1 + m_2)v_1^2$. By conservation of energy $m_1gh=\frac{1}{2}(m_1 + m_2)v_1^2$, so $v_1=\sqrt{\frac{2m_1gh}{m_1 + m_2}}$. Since we know $v_1 = 5.23$ m/s, we can assume the values of $m_1$, $m_2$, and $h$ were used to calculate it.

Step2: Analyze the motion of the crewman after the crate hits the ground

After the crate hits the ground, the crewman is in free - fall. The initial velocity of the crewman for this free - fall motion is $v_1$. The height the crewman falls is $H - h$. Using the kinematic equation $v_2^2=v_1^2+2g(H - h)$.
Let's assume $H - h$ is known. If we use the kinematic equation $v_2^2=v_1^2 + 2g\Delta y$, where $\Delta y=H - h$. Given $v_1 = 5.23$ m/s and $g = 9.8$ m/s², we have $v_2=\sqrt{v_1^2+2g(H - h)}$.

Answer:

To find the correct $v_2$, we need to know the values of $v_1$, $g$, and $H - h$. If we assume $v_1 = 5.23$ m/s, $g = 9.8$ m/s² and say $H - h = 10$ m for example:
\[v_2=\sqrt{5.23^2+2\times9.8\times10}=\sqrt{27.3529 + 196}=\sqrt{223.3529}\approx14.94\] m/s. Without specific values of $H$ and $h$, we can't give a definite answer. But the formula to calculate $v_2$ is $v_2=\sqrt{v_1^2+2g(H - h)}$ where $v_1$ is the speed of the crate when it hits the ground, $g$ is the acceleration due to gravity, and $H - h$ is the height the crewman falls after the crate hits the ground.