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3 years ago

What is the restoring force of a spring stretched 0.35 meters with a spring contact of 55 newtons?

Physics

2 answers:

Vesna [10]3 years ago

6 0

F=-kx = -19,25 answer D

Bogdan [553]3 years ago

6 0

Answer:

F=19.25 N

Explanation:

The restoring force for a spring is given by:

$\vec{F}=-k \vec{x}$

Where k is the spring constant and $\vec{x}$ is the spring deformation.

The minus sign indicates that force is opposed to the deformation direction.

According to the problem:

$k=55 \;N/m$ and $x=0.35 \;m$

Thus, in magnitude, the restoring force is:

$F=55\frac{N}{m} *0.35m=19.25\; N$

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