A 21700 kg train is reaching the end of the line. In order to slow down, the front of the train snags a cable that is attached to a spring. If the spring has a spring constant of 73400 N/m and the train's initial speed is 53.0 m/s, how much will the spring stretch to stop the train?
To find the amount of stretch in the spring, we can use the principle of conservation of mechanical energy.
The initial mechanical energy of the train is given by the kinetic energy:
Initial kinetic energy = (1/2) * mass * velocity^2
The final mechanical energy of the train is given by the potential energy stored in the stretched spring:
Final potential energy = (1/2) * spring constant * stretch^2
According to the principle of conservation of mechanical energy, the initial kinetic energy must be equal to the final potential energy.
Step 1: Calculate the initial kinetic energy:
Initial kinetic energy = (1/2) * mass * velocity^2
= (1/2) * 21700 kg * (53.0 m/s)^2
Step 2: Calculate the stretch using the potential energy equation:
Stretch = sqrt((2 * Initial kinetic energy) / spring constant)
= sqrt((2 * Initial kinetic energy) / 73400 N/m)
Step 3: Substitute the values into the equation and solve for the stretch:
Stretch = sqrt((2 * (1/2) * 21700 kg * (53.0 m/s)^2) / 73400 N/m)
Once you plug in the values and perform the calculations, you will find the stretch in the spring that stops the train.
To find the spring stretch, we can use the principle of conservation of mechanical energy.
The initial kinetic energy of the train is given by KE_initial = 0.5 * mass * velocity^2.
The final potential energy stored in the spring is given by PE_final = 0.5 * k * stretch^2, where k is the spring constant and stretch is the spring stretch.
Since the train is being slowed down, the kinetic energy is converted to potential energy in the spring. Therefore, we have:
KE_initial = PE_final
Substituting the values:
0.5 * mass * velocity^2 = 0.5 * k * stretch^2
Simplifying:
0.5 * 21700 kg * (53.0 m/s)^2 = 0.5 * 73400 N/m * stretch^2
Solving for the spring stretch:
stretch^2 = (0.5 * 21700 kg * (53.0 m/s)^2) / (0.5 * 73400 N/m)
stretch^2 = (0.5 * 21700 kg * 2809 m^2/s^2) / (0.5 * 73400 N/m)
stretch^2 = 311,765,000 kg * m^2 / N
stretch = √(311,765,000 kg * m^2 / N)
Calculating the spring stretch:
stretch ≈ 8,966.6 m
Therefore, the spring will stretch approximately 8,966.6 meters to stop the train.