A1.50 kg object is held 1.27 m above a relaxed, massless vertical spring with a force constant of 300 n/m. the object is dropped onto the spring. (a) how far does the object compress the spring?

0.353 m

Explanation:

According to the law of conservation of energy, the initial energy that the object possesses (in the form of gravitational potential energy) is converted into elastic potential energy of the spring.

The initial gravitational potential energy of the object is:

where

m = 1.50 kg is the mass of the object

g = 9.8 m/s^2 is the acceleration of gravity

h = 1.27 m is the height of the object

The final elastic potential energy of the compressed spring is

where

k = 300 N/m is the spring constant

x is the compression of the spring

Equalizing the two energies, we find:

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