An elevator cable breaks when a 920 kg elevator is 23.0 m above a huge spring (k

Question

An elevator cable breaks when a 920 kg elevator is 23.0 m above a huge spring (k = 2.10×105 N/m ) at the bottom of the shaft.

Part A

Calculate the work done by gravity on the elevator before it hits the spring.

Express your answer using three significant figures.

Part B

Calculate the speed of the elevator just before striking the spring.

Express your answer using three significant figures.

Part C

Calculate the magnitude of the amount the spring compresses (note that work is done by both the spring and gravity in this part).

Express your answer using three significant figures.

Explanation / Answer

a) Work done by weight on elevator prior to hitting spring = mgh
mgh = (920)(9.81)(23) =207579.6 J

b) V = 2gh = [(2)(9.81)(23)] = 451.26 = 21.24 m/s

c) Work done by weight is stored as spring P.E. and further work by gravity:
207579.6 = 1/2kx² +mgx {where k = 2.10×105 N/m, x = compression distance}
0 = -207579.6 + (0.5)(2.10×105)x² + (920)(9.81)x
0 = 1.05×105 x² + 9025.2 x - 207579.6
solve for x, positive root = 1.36 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.