a strechy material with a spring constant of 68.7 n/m is stretched 2.71 cm from its equilibrium length. what is the magnitude of the resulting spring force?



Answer :

The magnitude of the resulting spring force when the material is stretched from its equilibrium position is -0.02522 N.

The spring force is a spring or a stretchy material is given by the formula,

F = -1/2Kx²

Where z

F is the spring force,

K is the spring constant,

x is the compression or extension in the material,

The negative sign represent that the force is restoring in nature.

We know, in our case,

The extension in the material is 2.71cm.

1 cm = 1/100m

So,

2.71 cm = 2.71/100m

The spring constant is 68.7N/m.

Now, putting all the values,

F = 1/2Kx²

F = -1/2(68.6)(2.71/100)²

F = -0.02522N.

So, the spring force is 0.02522N.

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