Question

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. Spring coefficient is 10.6 kN/m. During the motion a safety clamp applies a constant 17000-N frictional force to the elevator.What is the speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring?

Answer 1

Answer:

v=3.649 m/s

Explanation:

Lets start with the force of gravity on the elevator.

$Fg=2000 kg* 9.8\frac{m}{s^{2} } \\Fg= 19600N$

But the friction clamp opposes this with a force of 17000 N

So the Net force on the elevator is

$Ft=19600 - 17000 \\Ft= 2600 N$

Kinetic Energy

$K=\frac{1}{2}*m*v^{2}\\K=\frac{1}{2}*2000kg*(4\frac{m}{s}) ^{2}\\K=16000J$

The motion will be describe

original Kinetic energy + work done = final kinetic energy + spring energy

$Ek+Ft=Ekf+Fk\\16000J+2600J=\frac{1}{2}*m*v^{2}+\frac{1}{2}*k*x^{2} \\18600J=\frac{1}{2}*2000kg*v^{2}+ \frac{1}{2}*10.6x10^{3}\frac{N}{m} *1m^{2}\\18600-5300=\frac{1}{2}*2000kg*v^{2}\\v^{2}=\frac{13300J}{1000kg}\\v^{2}=13.3\\v=\sqrt{13.3}=3.64 \frac{m}{s}$