A block of mass 2.5kg is pressed against a spring and compresses the spring to a length of 10.0 cm. Then, the block is released and begins to move to the right as the spring returns
to its natural length of 21.4 cm. The spring has a spring constant of 88 N/m. Using Hooke's
Law, determine the force in the spring.
Pay attention to your units (specifically your values of spring length). Answer in units of

10.032 N.

Explanation:

From the question given above, the following data were obtainedb

Mass (m) = 2.5 Kg

Final length = 10.0 cm

Original length = 21.4 cm

Spring constant (K) = 88 N/m

Force (N) =?

Next, we shall determine the compression of the spring. This can be obtained as follow:

Final length = 10.0 cm

Original length = 21.4 cm

Compression (e) =?

e = Original length – final length

e = 21.4 – 10

e = 11.4 cm

Next, we shall convert 11.4 cm to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

11.4cm = 11.4 cm × 1 m / 100 cm

10 cm = 0.114 m

Finally, we shall determine the force. This can be obtained as illustrated below:

Compression (e) = 0.114 m

Spring constant (K) = 88 N/m

Force (N) =?

F = Ke

F = 88 × 0.114

F = 10.032 N

Thus, the force in the spring is 10.032 N

their crystal patterns have been disrupted by stress