Physics, 17.10.2019 18:40 pingkeara

In a "worst-case" design scenario, a 2000-kg elevator with broken cables is falling at 4.00 m/s when it first contacts a cushioning spring at the bottom of the shaft. the spring is supposed to stop the elevator, compressing 2.00 m as it does so. during the motion a safety clamp applies a constant 17000-n frictional force to the elevator.

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The speed of the elevator after covering after the point of its contact with the spring will be $\boxed{3.65\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The mass of the elevator is $2000\,{\text{Kg}$ .

The speed of the elevator after the cable breaks is $4\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}$ .

The friction force by the safety clamp is $17000\,{\text{N}$ .

Concept:

As the cable of the elevator breaks, the elevator starts to fall with some velocity and the friction provided by the cushion spring will start to decrease the speed of the elevator.

The total amount of kinetic energy of the elevator as it starts moving is:

$\begin{aligned}{K_i}&= \frac{1}{2}m{v^2}\\&=\frac{1}{2}\times 2000 \times {\left( 4 \right)^2}\\&= 16000\,{\text{J}}\\\end{aligned}$

The net force acting on the elevator during its drag of over the safety clamp is:

$\begin{aligned}{F_{net}} &= {F_g} - f\\&= mg - f\\&= \left( {2000 \times 9.8} \right) - 17000\\&= 2600\,{\text{N}}\\\end{aligned}$

Finally when the elevator covers distance, the elevator has some kinetic energy due to its speed and rest of the energy is exhausted in the work done by the net force in covering the next over the spring.

The work done by the force in covering $1\,{\text{m}}$ is:

$\begin{aligned}{W_i}&= {F_{net}} \times d\\&= 2600 \times 1\\&= 2600\,{\text{J}}\\\end{aligned}$

Apply the conservation of energy for the elevator.

$\begin{aligned}{K_i}&= {K_f} + {W_i}\\16000&= \frac{1}{2}mv_f^2 + 2600\\{v_f}&=\sqrt{\frac{{2 \times 13400}}{{2000}}}\\ &=3.66\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}} {\text{s}}}}\right.\kern\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

Thus, the speed of the elevator after covering after the point of its contact with the spring will be $\boxed{3.65\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

Learn More:

1. You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves down a frictionless track

2. It's been a great day of new, frictionless snow. Julie starts at the top of the 60∘ slope shown

3. The amount of kinetic energy an object has depends on its

Answer Details:

Grade: College

Subject: Physics

Chapter: Conservation of Energy

Keywords: Elevator, worst-case, design scenario, broken cables, contacts a cushioning spring, stop the elevator, conservation of energy, initial kinetic energy, work done.

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In a "worst-case" design scenario, a 2000-kg elevator with broken cables is falling at 4.00 m/s when...

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