A horizontal spring with spring constant 750 N/m is attached to a wall. An athlete presses against the free end of the spring, compressing it 5.0 cm. How hard is the athlete pushing?

Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.

Using the expression for hook's law,

F = ke.............. Equation 1

F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.

Given: k = 750 N/m, e = 5.0 cm = 0.05 m

Substitute into equation 1

F = 750(0.05)

F = 37.5 N

Hence the athlete is pushing 37.5 N hard

stigawithfun stigawithfun · Answer 2 · 2020-04-05T19:38:45+02:00

Answer:

37.5N

Explanation:

According to Hooke's law, the load or force, F, applied on an elastic material (e.g a spring) is directly proportional to the extension or compression, e, caused by the load. i.e

F ∝ e

F = k x e -------------------------(i)

where;

k = proportionality constant known as the spring constant.

From the question;

k = 750N/m

e = 5.0cm = 0.05m

Substitute these values into equation (i) as follows;

F = 750 x 0.05

F = 37.5N

Therefore, the load applied which is a measure of how hard the athlete is pushing is 37.5N