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?

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Physics - bobpursley, Wednesday, April 4, 2007 at 9:07pm
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

Bob, are we solving for h+x or only h. Because solving for only h gives a wrong answer

h+x is the initial height above the COMPRESSED spring, which is what they ask for

Thanks for the clarification

To solve this problem, we need to apply the principle of conservation of mechanical energy. Initially, the block is at a certain height above the compressed spring, and when it falls it will have both gravitational potential energy and elastic potential energy stored in the compressed spring.

First, let's calculate the gravitational potential energy of the block when it is at its initial height h above the compressed spring. The formula for gravitational potential energy is given by:

PE_g = m * g * h,

where m is the mass of the block (0.30 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the initial height above the compressed spring. So, the gravitational potential energy is:

PE_g = 0.30 kg * 9.8 m/s^2 * h.

Next, let's calculate the elastic potential energy stored in the compressed spring. The formula for elastic potential energy is given by:

PE_s = (1/2) * k * x^2,

where k is the spring constant (450 N/m) and x is the compression of the spring (3.0 cm = 0.03 m). So, the elastic potential energy is:

PE_s = (1/2) * 450 N/m * (0.03 m)^2.

According to the conservation of mechanical energy, the initial gravitational potential energy of the block when it is at height h above the compressed spring should be equal to the total elastic potential energy stored in the spring when the block is momentarily stopped.

Setting up the equation:

PE_g = PE_s,

0.30 kg * 9.8 m/s^2 * h = (1/2) * 450 N/m * (0.03 m)^2.

Now, we can solve for h:

h = (1/2) * 450 N/m * (0.03 m)^2 / (0.30 kg * 9.8 m/s^2).

Calculating this expression, we find:

h ≈ 0.022 m = 2.2 cm.

Therefore, the block was dropped from a height of approximately 2.2 cm above the compressed spring.