00:01
In this question we've been told that an elevator with a mass of 1 ,500 kg is supported by a steel cable.
00:07
The cable breaks when the bottom of the elevator is 10 meter above a spring, which has a force constant of 10 ,000 newton per meter.
00:16
So in the first part of the question, we need to find the speed of the elevator when it first contacts the spring.
00:22
So we need to find the speed.
00:26
So we know that it is 10 meter above the spring.
00:30
So we know the distance we know that it will be under the force of gravity so we are going to use a formula 2 as equals to v square minus u square so v is the initial v is the final speed and u is the initial speed so we know that in this case the initial speed is going to be 0 so 2 as equals to v square and v equals to under root 2 s so this is going to be under root 2 multiplied by acceleration in this case will be gravitational acceleration which is 9 .8.
01:13
1 multiplied by the distance which is s 10 meters.
01:19
So v is going to be 14 meter per second.
01:26
So this is the velocity of the elevator when it first contact the spring.
01:32
And this is the answer for part a.
01:33
In part b, we need to find out by what distance is the spring compressed when the force of the spring is equal to the force of gravity...