A vertical spring with a spring constant of 410 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.26-kg block is dropped from rest. It collides with and sticks to the spring, which is compressed by 2.8 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height above the compressed spring was the block dropped?
To find the height from which the block was dropped, we need to apply the principles of conservation of energy.
First, we need to determine the potential energy of the block at the initial height before it is dropped. We can calculate this using the formula:
Potential Energy (PE) = mgh
Where:
m = mass of the block (0.26 kg)
g = gravitational acceleration (approximately 9.8 m/s²)
h = initial height
Next, we need to consider the energy at the point where the block is momentarily brought to a halt and the spring is compressed. At this point, the total energy of the system is converted from potential energy to elastic potential energy stored in the spring. The elastic potential energy of a spring can be calculated using the formula:
Elastic Potential Energy (PE_elastic) = (1/2)kx²
Where:
k = spring constant (410 N/m)
x = compression of the spring (2.8 cm = 0.028 m)
Since the block comes to a momentary halt at the lowest point of its motion, all of its initial potential energy is converted into the elastic potential energy stored in the spring:
PE = PE_elastic
Using the equations mentioned above, we can set them equal to each other and solve for the initial height:
mgh = (1/2)kx²
Substituting the given values:
0.26 * 9.8 * h = (1/2) * 410 * 0.028²
Calculating this equation will give us the value of h, which is the initial height from which the block was dropped.