Question

1. an elastic climbing rope is tested by fixing one end of the rope to the top of a crane. the other end of the rope is connected to a block which is initially at position a. the block is released from rest. the mass of the rope is negligible. the unextended length of the rope is 60.0 m. from position a to position b, the block falls freely. a. at position b the rope starts to extend. calculate the speed of the block at position b. b. at position c the speed of the block reaches zero. the time taken for the block to fall between b and c is 0.759 s. the mass of the block is 80.0 kg. i. determine the magnitude of the average resultant force acting on the block between b and c.

Explanation:

Step1: Use free - fall motion formula

We use the equation $v^{2}=v_{0}^{2}+2as$. The block starts from rest ($v_{0} = 0$), the acceleration $a = g=9.8\ m/s^{2}$ and the distance $s = 60.0\ m$.
$v^{2}=0 + 2\times9.8\times60$

Step2: Solve for $v$

$v=\sqrt{2\times9.8\times60}=\sqrt{1176}\approx34.3\ m/s$

Step3: For part b(i), use impulse - momentum theorem

The impulse - momentum theorem is $F_{avg}\Delta t=\Delta p$. The initial velocity at B is $v_{B}\approx34.3\ m/s$ and the final velocity at C is $v_{C} = 0$. The mass of the block $m = 80.0\ kg$. $\Delta p=m(v_{C}-v_{B})=80\times(0 - 34.3)=- 2744\ kg\cdot m/s$. The time $\Delta t=0.759\ s$.
$F_{avg}=\frac{\Delta p}{\Delta t}=\frac{-2744}{0.759}\approx - 3615\ N$. The magnitude of the average resultant force is $|F_{avg}| = 3615\ N$

Answer:

a. $34.3\ m/s$
b. i. $3615\ N$