A vertical spring with a spring constant of 450 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30 kg block is dropped from rest. It collides with and sticks to the spring, which is compressed by 3.0 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in cm) above the compressed spring was the block dropped?
The spring compression had 1/2 k x^2 of work done on it. Assuming no losses to friction, then the energy that went into it was mg(h+x). Calculate h
Please clarify your response.
Thanks!
Sure! Let's break down the problem step by step.
First, let's analyze the energy transformation that occurs when the block is dropped onto the spring. Initially, the block has gravitational potential energy due to its height above the compressed spring. As it falls, this potential energy is converted to kinetic energy. When the block collides with the spring, some of the kinetic energy is transferred into the compression of the spring.
The equation for gravitational potential energy is given by U = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height above the compressed spring.
The equation for kinetic energy is given by K = (1/2)mv^2, where m is the mass of the block and v is its velocity.
Since the block starts from rest, it has no initial kinetic energy. Therefore, the total initial energy of the system is equal to the gravitational potential energy of the block:
U_initial = mgh
After the block collides and comes to a momentary halt, all the initial energy is converted into the compression of the spring. The work done on the spring is given by W = (1/2)kx^2, where k is the spring constant and x is the compression of the spring.
Therefore, the total final energy of the system is equal to the work done on the spring:
W_final = (1/2)kx^2
Since there is no loss of energy due to friction, we can equate the initial and final energies:
U_initial = W_final
mgh = (1/2)kx^2
Now, let's plug in the given values. The mass of the block is 0.30 kg, the spring constant is 450 N/m, and the compression of the spring is 3.0 cm (which is equivalent to 0.03 m):
(0.30 kg)(9.8 m/s^2)h = (1/2)(450 N/m)(0.03 m)^2
Simplifying the equation gives:
2.94h = 0.2025
Solving for h will give us the height above the compressed spring:
h = 0.2025 / 2.94 ≈ 0.0689 m
Finally, to convert the height to centimeters, we multiply by 100:
h ≈ 6.89 cm
Therefore, the block was dropped from a height of approximately 6.89 cm above the compressed spring.
I hope this clarifies the response! Let me know if you have any further questions.