An ore car of mass 40000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 16 m lower vertically, is a horizontally situated spring with constant 2.7 × 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.
MY ANSWER
Fs = Kx
x = K(Fs)
x = (2.7*10^5 N/m)(40,000 kg * 9.8m/s^2)
where
Vinitial = 0m/s
d = 16m
k = 2.7*10^5 N/m
a = 9.8m/s^2
No friction.
x = ? meters
I think they're hinting at energy:
mgh = 1/2 k x^2
To determine how much the spring compresses in stopping the ore car, we can use the principle of work and energy.
First, let's calculate the potential energy gained by the ore car as it rolls downhill. The potential energy is given by the formula:
PE = mgh
where m is the mass of the car (40,000 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical height difference (16 m). Substituting the given values, we have:
PE = (40,000 kg)(9.8 m/s^2)(16 m)
Next, we can calculate the spring potential energy at the end of the tracks. The spring potential energy is given by the formula:
PE = (1/2)kx^2
where k is the spring constant (2.7 × 10^5 N/m) and x is the compression of the spring. Since the car comes to rest, all the potential energy gained by the car is transferred to the spring potential energy. Therefore, we can equate the two potential energies:
(40,000 kg)(9.8 m/s^2)(16 m) = (1/2)(2.7 × 10^5 N/m)x^2
Now we can solve for x:
x^2 = 2(40,000 kg)(9.8 m/s^2)(16 m) / (2.7 × 10^5 N/m)
x^2 = 2(40,000 kg)(9.8 m/s^2)(16 m) / (2.7 × 10^5 N/m)
x^2 ≈ 589.63
Taking the square root of both sides, we get:
x ≈ √589.63
x ≈ 24.3 meters
Therefore, the spring is compressed by approximately 24.3 meters in stopping the ore car.