A vertical spring with a spring constant of 400 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 2.3 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?
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To solve this problem, we can use the principle of conservation of mechanical energy. The initial potential energy of the block when it is dropped is equal to the final potential energy of the block-spring system when the block is momentarily at rest.
The potential energy of the block when it is dropped is given by:
Initial potential energy = mgh
The potential energy of the block-spring system when the block is momentarily at rest is given by:
Final potential energy = (1/2)kx^2
Where:
m = mass of the block = 0.30 kg
g = acceleration due to gravity = 9.8 m/s^2
h = initial height above the compressed spring
k = spring constant = 400 N/m
x = compression of the spring = 2.3 cm = 0.023 m
Setting the initial and final potential energies equal, we have:
mgh = (1/2)kx^2
Solving for h:
h = (1/2)(kx^2) / mg
Substituting the given values:
h = (1/2)(400 N/m)(0.023 m)^2 / (0.30 kg)(9.8 m/s^2)
Calculating h:
h ≈ 0.002 m
Converting h to centimeters:
h ≈ 0.002 m * 100 cm/m ≈ 0.2 cm
Therefore, the block was dropped from a height of approximately 0.2 cm above the compressed spring.
To find the height from which the block was dropped above the compressed spring, we can use the concept of conservation of mechanical energy.
First, let's consider the initial potential energy (PE) of the block when it is dropped from rest. The initial potential energy can be calculated using the formula:
PE_initial = m * g * h_initial
where:
m = mass of the block = 0.30 kg
g = acceleration due to gravity = 9.8 m/s^2
h_initial = initial height
Next, let's consider the potential energy stored in the compressed spring. The potential energy stored in a spring can be calculated using the formula:
PE_spring = (1/2) * k * x^2
where:
k = spring constant = 400 N/m
x = compression of the spring = 2.3 cm = 0.023 m
According to the conservation of mechanical energy, the initial potential energy of the block must be equal to the potential energy stored in the compressed spring. So we can set up the equation:
m * g * h_initial = (1/2) * k * x^2
Now, we can rearrange the equation to solve for h_initial:
h_initial = (1/2) * k * x^2 / (m * g)
Substituting the given values, we get:
h_initial = (1/2) * 400 N/m * (0.023 m)^2 / (0.30 kg * 9.8 m/s^2)
Now we can calculate h_initial:
h_initial ≈ 0.0155 m
Finally, we convert the height from meters to centimeters:
h_initial ≈ 1.55 cm
Therefore, the block was dropped from a height of approximately 1.55 cm above the compressed spring.