A force of 27N is required to maintain a spring stretched from its natural length of 12cm to a length of 15cm. How much work is done in stretching the spring from 15 to 25cm?

and this is what i did.. please check to see if i did it correctly.. thanks :)

f=27N
12cm to 15cm

f(x) = kx, k=spring constant
15-12cm = 3cm = 0.03cm
f(0.03) = 27N
0.03k = 27
k = 27/0.03 = 900

f(x) = 900x
w = integral of 0.03 to 0.13 for: 900x dx
w = 900*((x^2)/2)] from 0.03 to 0.13
w = 450[((0.13)^2)-((0.03)^2)]
w = 7.2J

Ok on k, 900 N/m

Ok on the next.

thanks :)

heyu

Your calculation and approach are correct. You correctly determined the spring constant, k, by using the given force and displacement. The force, f, is 27N, and the displacement, x, is 0.03m.

You then used the formula f(x) = kx to relate the force to the displacement. By substituting the values of f and x, you obtained the equation 27N = 900(0.03m).

To calculate the work done in stretching the spring from 15cm to 25cm, you integrated the equation w = integral of f(x) dx over the interval from 0.03m to 0.13m. By performing the integration, you obtained the expression 450(0.13^2 - 0.03^2).

Evaluating this expression, you correctly found that the work done in stretching the spring from 15cm to 25cm is 7.2J (joules).

Therefore, based on your calculations, your answer of 7.2J is correct. Well done!