A block with mass 0.470 kg sits at rest on a light but not long vertical spring that has spring constant 85.0 N/m and one end on the floor. A second identical block is dropped onto the first from a height of 4.40 m above the first block and sticks to it. What is the maximum elastic potential energy stored in the spring during the motion of the blocks after the collision?

The maximum potential energy stored in the spring during the motion of the blocks is 20.39 J.

The given parameters;

The extension of the spring is calculated as follows;

The maximum potential energy stored in the spring during the motion of the blocks is calculated as;

Thus, the maximum potential energy stored in the spring during the motion of the blocks is 20.39 J.

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elastic potential energy = 20.27 J

Explanation:

Given that the

Mass M = 0.470 kg

Height h = 4.40 m

Spring constant K = 85 N/m

The maximum elastic potential will be equal to the maximum kinetic energy experienced by the block.

But according to conservative of energy, the maximum kinetic energy is equal to the maximum potential energy experienced by the block of mass M.

That is

K .E = P.E = mgh

Where g = 9.8m/s^2

Substitutes all the parameters into the formula

K.E = 0.470 × 9.8 × 4.4

K.E = 20.27 J

Where K.E = maximum elastic potential energy stored in the spring during the motion of the blocks after the collision which is 20.27J.

e) none of the above answers are correct