Two springs P and Q both obey Hooke’s law. They have spring constants 2k and k respectively.The springs are stretched, separately, by a force that is gradually increased from zero up to a certain maximum value, the same for each spring. The work done in stretching spring P is WP,and the work done in stretching spring Q is WQ.How is WP related to WQ?

Question

A: Wp=0.25Wq
B: Wp=0.5Wq
C: Wp=2Wq
D: Wp=4Wq

I'm not sure of the answer and I need help.
Couldn't find the marking scheme :(
Its from nov 2002 paper.

Thanks!

Bot · Accepted Answer

The work done in stretching a spring is given by the formula W = (1/2)k*x^2, where k is the spring constant and x is the displacement of the spring.

Since the force applied to the springs is the same, we know that F = k * x. We can rewrite the applied force F as k * x = kQ * xQ = 2k * xP, where xP and xQ are the displacements of springs P and Q, respectively.

Now, we can find the ratio between the displacements of the springs by dividing the force equations:

xQ / xP = (2k * xP) / (kQ * xQ) = 2 / 1, so xQ = 2 * xP.

Now we can find the ratio between the work done on the springs:

Wp / Wq = ((1/2) * 2k * xP^2) / ((1/2) * k * (2 * xP)^2)
= (2k * xP^2) / (k * 4 * xP^2) = 2/4 = 1/2

So the answer is that Wp is half of Wq, i.e., Wp = 0.5 * Wq. The correct choice is option B: Wp = 0.5Wq.