An ore car of mass 42000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 6.8 m lower vertically, is a horizontally situated spring with constant 3.8 x 10^5 N/m. The acceleration of gravity is 9.8 m/s^2.
How much is the spring compressed in stoping the ore car?
nbm
To find how much the spring is compressed in stopping the ore car, we can use the principle of conservation of mechanical energy.
1. First, let's calculate the potential energy of the ore car at the top of the incline. This potential energy will be converted into the potential energy of the compressed spring at the bottom of the incline.
Potential energy at the top (PE_top) = m * g * h_top
where m is the mass of the ore car, g is the acceleration due to gravity, and h_top is the vertical height from the top of the incline to the horizontal position of the spring.
In this case, the vertical height h_top is given as 6.8 m.
2. Next, let's calculate the potential energy stored in the compressed spring at the bottom (PE_spring).
Potential energy stored in spring (PE_spring) = (1/2) * k * x^2
where k is the spring constant and x is the compression of the spring. In this case, the spring constant k is given as 3.8 x 10^5 N/m.
3. According to the conservation of mechanical energy, the potential energy at the top of the incline must be equal to the potential energy stored in the compressed spring at the bottom. So we set up the following equation:
PE_top = PE_spring
m * g * h_top = (1/2) * k * x^2
4. Rearrange the equation to solve for x:
x^2 = (2 * m * g * h_top) / k
x = sqrt((2 * m * g * h_top) / k)
Now we can substitute the given values into the equation to find x, which represents the compression of the spring in stopping the ore car.