spring problem

[kpx001]

It requires 10 in-punds of work to stretch a spring from its natural length of 10 inches to a length of 12 inches. How much additional work (in inch-pounds) is done in streching it one inch further.

ok this is how i approached this problem.
F=kx
10 = k(2) => k=5

intergral from 0 to 1 5x = [5x^2/2] => 5/2 ? answer is 25/2.

could someone explain what exaclty im doing because this is also a physics problem and i dont get the aditional one inch further. it wouldnt be 0 to 3 because the 0 to 2 constant is 5? im unsure but can someone make this easier? thank you

 

[Deleted member 4993]

It requires 10 in-punds of work to stretch a spring from its natural length of 10 inches to a length of 12 inches. How much additional work (in inch-pounds) is done in streching it one inch further.

ok this is how i approached this problem.
F=kx
10 = k(2) => k=5

intergral from 0 to 1 5x = [5x^2/2] => 5/2 ? answer is 25/2.

could someone explain what exaclty im doing because this is also a physics problem and i dont get the aditional one inch further. it wouldnt be 0 to 3 because the 0 to 2 constant is 5? im unsure but can someone make this easier? thank you

It should be

\(\displaystyle \int^{3}_{2}5x\, dx\)

 

[bertus]

It requires 10 in-punds of work to stretch a spring from its natural length of 10 inches to a length of 12 inches. How much additional work (in inch-pounds) is done in streching it one inch further.

Seems some confusions here. First work is measured in Joules it is a form or amount of energy. Force is measured in Newtons and work is force times distance. So I have to assume it 10 Joules of work to move it from 10 to 12 inches. I dont think there is a unit called in-punds.

Now there is an integral:
10 to 12 int(f(x)) dx = 10. But f(x) = kx where k is a constant. 1/2 kx ^ 2 10 to 12 = 10 So 1/2 k * 144 - 1/2 k * 100 = 10
22K = 10 K = 5/11. Now knowing K take the integral
12 to 13 int(5/11)x dx
Been long time but seem i did this correct?

 

[chivox]

It requires 10 in-punds of work to stretch a spring from its natural length of 10 inches to a length of 12 inches. How much additional work (in inch-pounds) is done in streching it one inch further.

Seems some confusions here. First work is measured in Joules it is a form or amount of energy. Force is measured in Newtons and work is force times distance. So I have to assume it 10 Joules of work to move it from 10 to 12 inches. I dont think there is a unit called in-punds.

Now there is an integral:
10 to 12 int(f(x)) dx = 10. But f(x) = kx where k is a constant.
...

The "in-punds" thing was a typo, I'm pretty sure. Poster meant inch-pounds, which is a valid unit of work. Also, 10 inches is the "natural" length of the spring, according to the original question. Doesn't that mean 10 inch-pounds of work were done to go from 0 to 2 inches?

kpx: The "k" is a property of the spring, so as bertus suggested, first find k using the work given in the problem for stretching the spring from 0 inches to 2 inches, and then compute the work using the integral given by Subhotosh. The value of "k" is a constant based on the spring itself. It doesn't change from one length to the next, or from 0 to 2 vs. 0 to 3 vs. 2 to 3 inches. It stays constant, because it describes the spring itself. A different spring would have a different value for k, though.