- #1

henry3369

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## Homework Statement

A 2000-kg elevator with broken cables in a test rig is falling at when it contacts a cushioning spring at the bottom of the shaft. The spring is intended to stop the elevator, compressing 2.00 m as it does so. During the motion a safety clamp applies a constant 17,000-N frictional force to the elevator. What is the necessary force constant k for the spring?

## Homework Equations

W

_{other}+ K

_{i}+U

_{i}= K

_{f}+U

_{f}

## The Attempt at a Solution

W

_{other}+ K

_{i}= U

_{f}

(-17000 * 2) + (1/2)(2000)(4

^{2}) = (1/2)k(2

^{2}) + (2000)(-9.8)(-2)

So the correct answer to the problem is k = 10600 N/m. My book solves this by making the final potential energy negative so that it is (2000)(9.8)(-2) and then solving for the spring constant. What I don't understand is why the force of gravity is not negative to make the entire final potential energy positive. The negative 2 clearly states that down is negative so why is gravity not negative if it is also pointing down?