Question

using a simple pulley and rope system, a crewman on an arctic expedition is trying to lower a crate of mass $m_1 = 5.39$ kg to the bottom of a steep ravine of height $h = 23.8$ m. the crewman of mass $m_2 = 50.3$ kg is walking while holding the rope, being careful to lower the crate at a constant speed of $v_0 = 1.50$ m/s. 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 $v_1$ will the crate hit the ground? $v_1 = 5.23$ m/s

Explanation:

Step1: Identify the system and forces

The system consists of the crate of mass $m_1 = 5.39$ kg and the crew - man of mass $m_2=50.3$ kg. The net - force on the system when the crewman slips is due to the weight of the crate. According to Newton's second law, $F_{net}=ma$. The net force $F_{net}=m_1g$, and the total mass of the system is $m = m_1 + m_2$. So, the acceleration $a=\frac{m_1g}{m_1 + m_2}$, where $g = 9.8$ m/s².
$a=\frac{5.39\times9.8}{5.39 + 50.3}=\frac{52.822}{55.69}\approx0.95$ m/s²

Step2: Use the kinematic equation

We know the initial velocity $v_0 = 1.50$ m/s, the displacement $d=h = 13.2$ m, and the acceleration $a$. We use the kinematic equation $v_1^{2}=v_0^{2}+2ad$.
Substitute the values: $v_1^{2}=(1.50)^{2}+2\times0.95\times13.2$.
$v_1^{2}=2.25+2\times0.95\times13.2=2.25 + 25.08=27.33$.
$v_1=\sqrt{27.33}\approx5.23$ m/s

Answer:

$5.23$ m/s