Question

Constants Part A A 2045 kg elevator with broken cables is falling at 4.60 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing by 2.80 m as it stops. During the motion a safety clamp applies a constant 1.64x104 N_frictional force to the elevator What is the spring constant of the spring that stops the elevator? Value Units Request Answer Submit Part B What is the speed of the elevator after it has moved downward 1.53 m from the point where it first contacts a spring? HA Value Units Request Answer Submit

Answer 1

This problem is from work and energy.We will use energy conservation and work energy theorem.

This problem is from work and energy. We will use energy conservation and work energy theorem. {Initial kinetic Energy + Initial potential energy} = {Energy stored in spring} + {loss due to friction } 1/2 (2.45) (4.6)^2 + (2045) (9.8) (2.8) = 1/2 k(2.8)^2 + (1.64 times 10^4) (2.8) k = 8120 N/m k = 8.12 times 10^3 N/m B) Net work done = change in kinetic energy {(2045) (9.8) (1.53) - (1.64 times 10^40 (2.8) - (0.5) (8120) (1.53)^2} = 1/2 mv^2 - 1/2 m(4.6)^2 m = 2045 kg v = 4.16 m/s