A work W0 is required to stretch a certain spring 2 cm from its equilibrium position.?

Question

What is the amount of work, W0, required to stretch a spring 2 cm from its equilibrium position? Additionally, could you provide the formula used to calculate this work, and any relevant information about the spring constant or other factors that may influence the calculation?

Anonymous · Accepted Answer

How much work is required to stretch the spring 1 cm from equilibrium?

Work = ∫F∙dx = ∫kx∙dx = ½kx² —–> Wo = ½k*0.02² = k/5000 so now we know k = 5000*Wo

½kx² = 2500Wox² —–> 2500Wo*0.01² = 0.25Wo = Wo/4 <——

½kx² evaluated from 0.02 to 0.03 = 2500*Wo(0.03² – 0.02²) = 1.25Wo = 5Wo/4

Anonymous · Answer

Work is equal to the change of elastic potential energy.
That’s a quadratic function of stretch length (x^2): W = 0.5kx^2
If you stretch it twice as much, required work quadruples (2^2 = 4)
If you stretch it half as much, it’s one quarter. Etc.
W = W0/4 for 1 cm.
Suppose the spring is already stretched 2 cm from equilibrium. How much additional work is required to stretch it to 3 cm from equilibrium?
If spring has stiffness k, then elastic potential energy at x=2 cm is W0 = 0.5*k*x^2 = 0.5 * k * 2^2 = 2k
(you can turn cm to m or you don’t have to, because it wil cancel anyway)
Now imagine we start here and stretch it even more to x=3 cm. We are doing work to increase the potential energy stored in it.
After we’re done, new total energy = 0.5 * k * x^2 = 4.5k
The change from x=2 to x=3 was 4.5k – 2k = 2.5 k
We already said W0 was 2k.
To find ratio, divide 2.5k / 2k = 1.25
So from x=2 to x=3 is 1.25W0.

Beatrice Berge · Answer

Yes, yes it is.

Anonymous

A work W0 is required to stretch a certain spring 2 cm from its equilibrium position.?

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