Problem 7.69 An elevator cable breaks when a 900 kg elevator is 22.0 m above the
ID: 1491902 • Letter: P
Question
Problem 7.69 An elevator cable breaks when a 900 kg elevator is 22.0 m above the top of a huge spring (k = 9.00×104 N/m ) at the bottom of the shaft. Part A Calculate the work done by gravity on the elevator before it hits the spring. WG = 1.94×105 J SubmitMy AnswersGive Up Correct Part B Calculate the speed of the elevator just before striking the spring. v = 20.8 m/s SubmitMy AnswersGive Up Correct Part C Calculate the amount the spring compresses (note that here work is done by both the spring and gravity). x = I need help with part c. I put mg=kx (900)(9.8)=9x10^4(x) and got x=0.098. this answer is incorrect
Explanation / Answer
(a) The work done by gravity is the decrease in the potential energy:
Wgrav = -mg(hf - hi) = -(900kg)(9.80m/s2)(0 - 22.0m) = 1.94 x 105J.
(b) The work done by gravity increases the kinetic energy: Wgrav = DK;
1.94 x 105J = 1/2(900kg)v2 - 0, which gives v = 20.8 m/s.
(c) For the motion from the break point to the maximum compression of the spring, we have
Wspring + Wgrav = DK;
-(1/2kxf2 - 1/2kxi2) - mg(hf - hi) = 1/2mvf2 - 1/2mvi2;
- [1/2(9.00 x 104N/m)x2 - 0] - (900kg)(9.80m/s2)(-x - 22.0m) = 0 - 0.
-45000 x2 + 8820 x + 194040 = 0
This is a quadratic equation for x, which has the solution x = -1.98 m, 2.18 m.
Because x must be positive, the spring compresses 2.18 m.