A large spring is placed at the bottom of an elevator shaft to minimize the impact in case the elevator cable breaks. A loaded car has mass 480 , and its maximum height above the spring is 11.8 . In order to minimize the shock, the maximum acceleration of the car after hitting the spring is 4. solve for the spring constant k
You need to include units with your numbers. That is one the first things you should have learned about physics.
We can solve for the spring constant, k, using the principle of conservation of energy.
Let's break down the problem step by step:
1. The maximum height above the spring is given as 11.8 units. This means that the potential energy of the loaded car at this height will convert into the elastic potential energy of the spring when it hits.
2. The potential energy of an object at height h is given by the formula: PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (which is approximately 9.8 m/s^2), and h is the height.
Plugging in the values, we have:
PE = (480 kg) * (9.8 m/s^2) * (11.8 m)
= 56,505.6 J
3. When the loaded car hits the spring, it compresses and stores energy as elastic potential energy. According to Hooke's Law, the elastic potential energy in a spring is given by the formula:
PE = (1/2) k x^2
Where k is the spring constant and x is the displacement from the equilibrium position.
4. We are given the maximum acceleration of the car after hitting the spring as 4 m/s^2. Since acceleration is the second derivative of displacement with respect to time, we can relate it to the spring constant using the equation:
a = (k / m) x
Rearranging the equation, we have:
k = a * m / x
5. We can substitute the given values into the equation:
k = (4 m/s^2) * (480 kg) / x
6. To calculate the displacement, x, we need to relate the potential energy to the elastic potential energy using the law of conservation of energy.
PE = (1/2) k x^2
Rearranging the equation, we have:
x = √(2 * PE / k)
7. Substituting the values we calculated earlier:
x = √(2 * 56,505.6 J / k)
8. Plugging this value of x back into the equation for k:
k = (4 m/s^2) * (480 kg) / (√(2 * 56,505.6 J / k))
Now, we need to solve this equation for k.
Let's simplify:
k = 1920 / (√(2 * 56,505.6 J / k))
Taking the square of both sides:
k^2 = (1920^2) / (2 * 56,505.6 J / k)
Simplifying further:
k^2 = 3600 k
k^2 - 3600 k = 0
Factoring out k, we have:
k (k - 3600) = 0
So, we have two solutions: k = 0 and k = 3600 N/m.
Since a spring constant of k = 0 doesn't make sense in this scenario, the spring constant k = 3600 N/m will minimize the shock and ensure the maximum acceleration of 4 m/s^2 is achieved after hitting the spring.