If an ore car with a mass of 38000 kg starts from rest & rolls downhill.....At the end of the tracks, 18 m lower vertically, is a horizontally situated spring with constant 520000 N/m. Acceleration of gravity is 9.8 m/s^2. Ignore friction.
How much is the spring compressed in stopping the ore car (answer in m)?
2006-10-09 08:05:30 · 2 answers · asked by Dee 4 in Science & Mathematics ➔ Physics
apply equation v^2 - u^2 =2as
v^2=2*9.8*18..............................(a)
gives u velocity after 18mtr
now u know
1/2kx^2=1/2m*v^2
condition for s.h.m
multily equqtion a by 1/2m
u will get ~ 6703200 N
equte it with eqution (b)
u will get x^2 as 257.81or x=(257)^1/2
2006-10-09 08:31:19 · answer #1 · answered by n nitant 3 · 0⤊ 0⤋
Easy one. 2.3984 speckometers. Ask a dumb question, get a dumb answer.
2006-10-09 15:11:05 · answer #2 · answered by 3810trebor@sbcglobal.net 2 · 0⤊ 1⤋