An ore car of mass 41000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 9.6 m lower vertically, is a horizontally situated spring with constant 4.7 × 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.
Gravitational potential energy loss = Spring potential energy increase
M g H = (1/2) k X^2
k = 4.7*10^5 N/m
M = 4.1*10^4 kg
H = 9.6 m
You known what g is.
Solve for X
Thank you!
I got it wrong. I did exactly what you said. Multiplied m times g times h then multiplied by 2. Sq root that number and got 2778.926411 but it was wrong.
X = sqrt(2 M g H/k)
= 4.54 m
You forgot to divide by k, the spring constant.
Well, I guess the ore car really sprung into action, huh? Let's calculate how much it compressed the spring!
First, we need to find the potential energy of the ore car when it starts rolling downhill. We can use the formula for gravitational potential energy: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height difference.
PE = (41000 kg)(9.8 m/s^2)(9.6 m) = 3.9 × 10^6 J
Now, let's find out how much the spring compressed. This energy will be transferred to the spring as potential energy. The formula for potential energy of a spring is: PE = (1/2)kx^2, where k is the spring constant and x is the displacement or compression.
3.9 × 10^6 J = (1/2)(4.7 × 10^5 N/m)x^2
Simplifying, we get:
7.8 = x^2
Taking the square root of both sides, we find:
x = √7.8 ≈ 2.79 m
So, the spring gets compressed by approximately 2.79 meters. I guess it really gets into the spring of things!
To solve this problem, we can use the principle of conservation of mechanical energy, which states that the initial mechanical energy of the system (ore car) is equal to the final mechanical energy.
The initial mechanical energy of the system consists only of gravitational potential energy (PE).
PE_initial = m * g * h_initial
where m is the mass of the ore car (41000 kg), g is the acceleration due to gravity (9.8 m/s^2), and h_initial is the initial height (0 m).
PE_initial = 41000 kg * 9.8 m/s^2 * 0 m = 0 J
The final mechanical energy of the system is the sum of the gravitational potential energy (PE_final) and the elastic potential energy stored in the spring (PE_spring).
PE_final = PE + PE_spring
PE_final = m * g * h_final + (1/2) * k * x^2
where h_final is the final height (9.6 m lower), k is the spring constant (4.7 × 10^5 N/m), and x is the amount the spring is compressed.
Substituting the given values:
PE_final = 41000 kg * 9.8 m/s^2 * (-9.6 m) + (1/2) * (4.7 × 10^5 N/m) * x^2
Since the ore car starts from rest, its initial velocity (v_initial) is 0 m/s. The final velocity (v_final) can be calculated using the equation:
v_final^2 = v_initial^2 + 2 * a * d
where a is the acceleration and d is the distance traveled.
Given that the ore car rolls downhill without friction, the acceleration (a) is equal to the acceleration due to gravity (g) and the distance traveled (d) is the vertical distance (9.6 m).
v_final^2 = 0^2 + 2 * 9.8 m/s^2 * 9.6 m
v_final^2 = 2 * 9.8 m/s^2 * 9.6 m
v_final^2 = 188.16 m^2/s^2
v_final = √188.16 m/s
The final kinetic energy (KE_final) can be calculated using:
KE_final = (1/2) * m * v_final^2
Substituting the given values:
KE_final = (1/2) * 41000 kg * (√188.16 m/s)^2
KE_final = (1/2) * 41000 kg * 13.697 m^2/s^2
KE_final = 281451.5 J
Since the initial mechanical energy is equal to the final mechanical energy:
0 J = m * g * h_final + (1/2) * k * x^2
0 = 41000 kg * 9.8 m/s^2 * (-9.6 m) + (1/2) * (4.7 × 10^5 N/m) * x^2
Now, we can solve for x:
0 = -3808080 N * m - (2.35 × 10^5 N/m) * x^2
Rearranging the equation:
(2.35 × 10^5 N/m) * x^2 = -3808080 N * m
Dividing both sides by (2.35 × 10^5 N/m):
x^2 = -3808080 N * m / (2.35 × 10^5 N/m)
x^2 = -16.2 m
Taking the square root of both sides:
x = √(-16.2) m
Since the square root of a negative number is undefined in the real number system, it indicates that the ore car does not compress the spring, or the compression is imaginary in this case. Therefore, the spring is not compressed in stopping the ore car.
Hence, the spring is not compressed.