An ore car of mass 36000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 21 m lower vertically, is a horizontally situated spring with constant5.1×10^5N/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
4567.8
To find the amount by which the spring is compressed in stopping the ore car, we can use the principle of conservation of mechanical energy.
First, let's calculate the potential energy of the ore car at the top of the hill and the potential energy converted into elastic potential energy in the spring at the bottom of the hill.
The potential energy (PE) of the ore car at the top of the hill is given by the product of its mass (m) and the gravitational acceleration (g) and the height (h) above the bottom of the hill:
PE_top = m * g * h
The potential energy converted into elastic potential energy in the spring at the bottom of the hill is given by the formula:
PE_spring = (1/2) * k * x^2
Where k is the spring constant and x is the distance by which the spring is compressed.
Since the total mechanical energy is conserved, the potential energy at the top of the hill is equal to the potential energy converted into elastic potential energy:
PE_top = PE_spring
So we have:
m * g * h = (1/2) * k * x^2
Rearranging the equation to solve for x, we get:
x = sqrt((2 * m * g * h) / k)
Plugging in the given values:
m = 36000 kg
h = 21 m
k = 5.1 × 10^5 N/m
g = 9.8 m/s^2
x = sqrt((2 * 36000 kg * 9.8 m/s^2 * 21 m) / (5.1 × 10^5 N/m))
Evaluating the expression, we get:
x ≈ 0.112 m
Therefore, the spring is compressed by approximately 0.112 meters in stopping the ore car.