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Predict the deformation or elongation of a spring that has a constant of elasticity of 400 N/m when a force of 75 N is applied in the rightward direction.

Respuesta :

Kazeemsodikisola

Answer:

Explanation:

Give that,

Spring constant (k)=40N/m

Force applied =75N

Since the force is applied to the right, we don't know if it is compressing or stretching the spring

So let assume it compress

Using hooke's law

F=-ke

e=-F/k

Then, e=-75/40

e=-1.875m

The deformation is 1.875m.

Let assume it stretch

Using hooke's law

-F=-ke

e=F/k

Then, e=75/40

e=1.875m

The elongation is 1.875m

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