Average power used to stretch a spring

[greg_rack]

Homework Statement

A light spring has unstretched length 0.40 m and spring constant 50 N/m .
The spring is stretched by a varying tension force that starts at a value of zero and increases at a constant rate of 0.20 N/s up to a maximum value.
When the force reaches its maximum value, the strain energy of the spring is 0.25 J. What is the average power used to stretch the spring?
(Assume that the spring obeys Hooke’s law.)

Relevant Equations

Strain energy
Hooke's law

First and foremost, I found the max stretch of the spring using the strain energy formula(x=√((2*0.25J)/k)) ).
Then, the maximum force exerted(Fmax=k*xmax), in order to find out the seconds needed for the force in [N/s] to reach its maximum value. Now, I got confused about how to find the AVERAGE POWER... and I thought about making the mean of all values assumed by the force in each second, with the formula(Pavg=0.2N*"sum of the first 'Δt' integers"/Δt). but with this unorthodox method, I have a result that isn't correct.

 

[hutchphd]

Your method should be approximately correct. But there is an easier way. You already figured the time (I presume correctly!)
Suppose I told you it took me 10 seconds to run 100 yds. How would you figure my average speed ?

 

[greg_rack]

Your method should be approximately correct. But there is an easier way. You already figured the time (I presume correctly!)
Suppose I told you it took me 10 seconds to run 100 yds. How would you figure my average speed ?

Well, so the avg power must be equal to ΔP/Δt, which is 25s. Indeed, its 0.25J/25s=0.01W, and its correct!
But I still have a question... why does de statement say the unstretched length of the spring? Isn't it unuseful?

 

[hutchphd]

Yes I think so. Of course figuring out what is useful is part of a quality education.
just for grins: how close to correct was your initial attempt? Do you see where the error comes in and how you would estimate it...there will be problems where that is the only approach...

 

[greg_rack]

Yes I think so. Of course figuring out what is useful is part of a quality education.
just for grins: how close to correct was your initial attempt? Do you see where the error comes in and how you would estimate it...there will be problems where that is the only approach...

I cannot really understand where does the big difference between 0.01W and 5.2e-3W(using the first method) comes from... I have calculated the power exerted in a second using the x(stretch)/s and then multiplied it by the sum of the first 25 numbers, and divided by 25... shouldn't it be at least approximately correct?

 

[haruspex]

I cannot really understand where does the big difference between 0.01W and 5.2e-3W(using the first method) comes from... I have calculated the power exerted in a second using the x(stretch)/s and then multiplied it by the sum of the first 25 numbers, and divided by 25... shouldn't it be at least approximately correct?

Your second method is not clear to me. Please explain in detail.

 

[greg_rack]

Your second method is not clear to me. Please explain in detail.

I started by calculating the stretching for the force acting in a second(0.2N), and then by finding the amount of energy needed to cause the same lengthening(U=0.5*k*x^2), ending up with an Energy/s(power). Then, by having the seconds during which the force acts, I made the mean of all the power values depending on the time: ((P*1s+P*2s+...+P*25s)/25=P(1+2+3+...25)/25).

 

[haruspex]

I started by calculating the stretching for the force acting in a second(0.2N), and then by finding the amount of energy needed to cause the same lengthening(U=0.5*k*x^2), ending up with an Energy/s(power). Then, by having the seconds during which the force acts, I made the mean of all the power values depending on the time: ((P*1s+P*2s+...+P*25s)/25=P(1+2+3+...25)/25).

Still not enough detail. I'll try to fill it in.
First second, extension reaches 4mm, Work done = 4e-4J, power=4e-4W.
2nd second, extension reaches 8mm, Work done = 12e-4J, power=12e-4W.
3rd second, extension reaches 12mm, Work done = 20e-4J, power=20e-4W.
:
25th second, extension reaches 100mm, Work done = 196e-4J, power=196e-4W.
Sum of powers=0.25W.
Avg power=0.01W.

So where did you go wrong?

 

[greg_rack]

Still not enough detail. I'll try to fill it in.
First second, extension reaches 4mm, Work done = 4e-4J, power=4e-4W.
2nd second, extension reaches 8mm, Work done = 12e-4J, power=12e-4W.
3rd second, extension reaches 12mm, Work done = 20e-4J, power=20e-4W.
:
25th second, extension reaches 100mm, Work done = 196e-4J, power=196e-4W.
Sum of powers=0.25W.
Avg power=0.01W.

So where did you go wrong?

A force of .2N causing stretching of 4mm, hasn't a work of 8e-4J(Work=FΔx)?
Anyway, I think the big mistake lies in the fact I've used the formula of strain energy(0.5*g*x^2) instead of the work, but why isn't it also correct?

 

[haruspex]

A force of .2N causing stretching of 4mm, hasn't a work of 8e-4J(Work=FΔx)?

It's only 0.2N at the end of 1s. On average it was only 0.1N. Work done on a spring is ½kx²=½(50)(0.004)²=4e-4.

the formula of strain energy(0.5*g*x^2)

Is g different from k?
If you'd really like to know where you went wrong, post all the details.

 

[greg_rack]

It's only 0.2N at the end of 1s. On average it was only 0.1N. Work done on a spring is ½kx²=½(50)(0.004)²=4e-4.

Is g different from k?
If you'd really like to know where you went wrong, post all the details.

I think I understood where the issue is... I assumed(incorrectly, I'd say) that each second the force exerted a power of P=½ (50)(0.004)²=4e-4W: then, with this assumption, power would have been P*1s on the 1st second, P*2s on the 2nd second and so on until P*25s. Dividing then the sum of all the values power has assumed during the 25s, by 25s, I would have get the mean power! Hope you got it now