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meikamae

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- Thread starter meikamae
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- #1

meikamae

- #2

schwarzchildradius

F = -kx

where k is the spring constant, F is the force generated by the spring, x is the displacement from equilibrium (where F=0). Any basic sample problem will require the equation re-arranged; or substitution of another variable into the two changable variables, x and F; or balance the equation with another force (say, a mass on a spring so that F = mg).

You could also ask to determine the velocity and KE of the spring at any time or displacement of x. Or you could find the general solution to the differential equation of a harmonic oscillator, which is what you've got with a mass on a spring, and find sinusoidal motion in space, decaying exponentially with the damping constant. So it depends on what depth you need.

- #3

HallsofIvy

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Your text may have F= -kx. The difference here is that F now is the force exerted BY the spring rather than the force exerted ON the spring ("equal and opposite").

Here are several "Hook's law" problems.

A spring with spring constant .4 cm/dyne has a force of 40 dynes applied to it (stretching it). How much does the spring stretch?

A force of 600 Newtons will compress a spring 0.5 meters. What is the spring constant of the spring?

A spring has spring constant 0.1 m/Newton. What force is necessary to stretch the spring by 2 meters?

A force of 40 Newtons will stretch a spring 0.1 meter. How far will a force of 80 Newtons stretch it?

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