An ore car of mass 44000 kg starts from rest
and rolls downhill on tracks from a mine. At
the end of the tracks, 26 m lower vertically, is
a horizontally situated spring with constant
4.5 × 105 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.
Eg=Ep
mgh=1/2kx^2
plug in givens and solve for x
To solve this problem, we need to apply the principle of conservation of mechanical energy. This principle states that the total mechanical energy of a system remains constant if there are no external forces doing work on the system. In this case, we can assume that there is no friction, so the only force that does work is the gravitational force.
The initial mechanical energy of the system is given by the potential energy of the ore car at the top of the hill, which is mgh, where m is the mass of the ore car, g is the acceleration due to gravity, and h is the height of the hill. In this case, mgh = (44000 kg)(9.8 m/s^2)(26 m) = 10,935,200 J.
When the ore car reaches the bottom of the hill, all of its potential energy has been converted into kinetic energy and the elastic potential energy of the compressed spring. The kinetic energy of the ore car is given by the formula (1/2)mv^2, where v is the velocity of the ore car. Since the ore car starts from rest, its initial velocity is 0, so the initial kinetic energy is also 0.
The final potential energy of the system is zero because the ore car is at the bottom of the hill. Therefore, the remaining energy is stored in the compressed spring. The elastic potential energy stored in a spring is given by the formula (1/2)kx^2, where k is the spring constant and x is the compression or extension of the spring.
Setting up the conservation of mechanical energy equation, we can write:
mgh = (1/2)mv^2 + (1/2)kx^2.
Since the velocity is not given directly, we can use the fact that the acceleration due to gravity is the same as the acceleration of the ore car as it rolls down the hill. Using the equation v^2 = u^2 + 2as, where u is the initial velocity (0 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and s is the vertical distance traveled, we can solve for v:
v^2 = (0 m/s)^2 + 2(-9.8 m/s^2)(26 m).
v^2 = 1016.8 m^2/s^2.
Taking the square root of both sides, v ≈ 31.898 m/s.
Substituting the values into the conservation of mechanical energy equation:
(44000 kg)(9.8 m/s^2)(26 m) = (1/2)(44000 kg)(31.898 m/s)^2 + (1/2)(4.5 × 10^5 N/m)x^2.
Adding the velocity term:
(44000 kg)(2548.8 J) = (1/2)(44000 kg)(1016.9516 J) + (1/2)(4.5 × 10^5 N/m)x^2.
Simplifying the equation:
112094720 J = 222480224.94 J + (2.25 × 10^11 N/m)x^2.
Rearranging the equation:
-110385504.94 J = (2.25 × 10^11 N/m)x^2.
Dividing both sides by (2.25 × 10^11 N/m):
-4.9056 × 10^-4 m = x^2.
Taking the square root of both sides:
x ≈ ±0.0221 m.
Since the problem asks for the compression of the spring, we take the positive value of x. Therefore, the spring is compressed by approximately 0.0221 meters in stopping the ore car.