An ore car of mass 41000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 29 m lower vertically, is a horizontally situated spring with constant 4.1 × 105 N/m.
The acceleration of gravity is 9.8 m/s2 . Ignore friction.
How much is the spring compressed in stop- ping the ore car?
Answer in units of m.
V^2 = Vo^2 + 2g*h = 0 + 19.6*29 = 568.4, V = 23.8 m/s.
Work = KE2-KE1 = KE2-0 = KE2.
Work = KE2 = 0.5M*V^2 = 0.5*41,000*(23.8)^2 = 11.7*10^6 J.
Work = F*d = F*29 = 11.7*10^6,
F = 4.03*10^5 N.
d = 1m/4.1*10^5N. * 4.03*10^5N. = = 0.983 m.
To find the compression of the spring, we need to calculate the potential energy gained by the ore car as it rolls downhill and compare it to the potential energy stored in the compressed spring.
1. Calculate the potential energy gained by the ore car:
Potential energy = mass * acceleration due to gravity * height
Potential energy = 41000 kg * 9.8 m/s^2 * 29 m
2. Calculate the potential energy stored in the compressed spring:
Potential energy = 0.5 * spring constant * spring compression^2
Now, we can set these two potential energies equal to each other:
41000 kg * 9.8 m/s^2 * 29 m = 0.5 * 4.1 × 10^5 N/m * spring compression^2
Rearranging the equation to solve for spring compression:
spring compression^2 = (41000 kg * 9.8 m/s^2 * 29 m) / (0.5 * 4.1 × 10^5 N/m)
spring compression^2 = 5707600 m^2
Finally, take the square root of both sides to find the spring compression:
spring compression = sqrt(5707600 m^2)
spring compression = 2389.62 m
Therefore, the spring is compressed by approximately 2389.62 meters in stopping the ore car.