An ore car of mass 37000 kg starts from rest
and rolls downhill on tracks from a mine. At
the end of the tracks, 9.9 m lower vertically,
is a horizontally situated spring with constant
3.9 × 10
5
N/m.
The acceleration of gravity is 9.8 m/s
2
.
Ignore friction.
How much is the spring compressed in stopping the ore car?
Answer in units of m
To find the compression of the spring when stopping the ore car, we can use the principle of conservation of mechanical energy. The initial potential energy of the ore car is converted into both kinetic energy and potential energy of the compressed spring.
First, let's find the potential energy of the ore car at the starting position:
Potential energy = mass * gravity * height
Potential energy = 37000 kg * 9.8 m/s^2 * 0 m (since it starts from rest)
Potential energy = 0 J
Next, let's find the kinetic energy of the ore car at the bottom of the tracks, just before it reaches the spring:
Kinetic energy = (1/2) * mass * velocity^2
Since the ore car started from rest, the initial velocity is 0 m/s.
Kinetic energy = (1/2) * 37000 kg * (0 m/s)^2
Kinetic energy = 0 J
The total mechanical energy at the bottom of the tracks is the sum of the kinetic energy and potential energy:
Total mechanical energy = 0 J + 0 J
Total mechanical energy = 0 J
At the end of the tracks, the potential energy is converted into the potential energy of the compressed spring:
Potential energy of the spring = (1/2) * spring constant * compression^2
0 J = (1/2) * 3.9 × 10^5 N/m * compression^2
Now, we can solve for the compression of the spring:
compression^2 = 0 J / [(1/2) * 3.9 × 10^5 N/m]
compression^2 = 0 m^2 / (1.95 × 10^5 N/m)
compression^2 = 0 m^2 / 1.95 × 10^5 kg/s^2
compression^2 = 0 m^2 / 1.95 × 10^5 m^2/s^2
compression^2 = 0 m^2 / 1.95 × 10^5 m^2/s^2
Since the numerator is zero, the compression of the spring is also zero.
Therefore, the spring is not compressed when stopping the ore car.