You pull on a spring, which obeys Hooke's law, in two steps. In step 1,you extend it by distance xs. In step 2, you further extend it by the same distance xs. Choose the answer that correctly compares the elastic potential energy of the spring after the first step (Us1) with that after the second step (Us2). Assume that the spring is initially undeformed.

A. Us2>2Us1 because the average force exerted on the spring in step 2 is larger than the average force exerted on the spring in step 1.
B. Us2=2Us1 because the force exerted on the spring at the end of step 2 is two times larger than at the end of step 1.
C. Us2>2Us1 because the more the spring is stretched, the larger the additional force that is needed to extend the spring for an additional unit distance.
D. Us2=2Us1 because the extension of the spring at the end of step 2 is two times larger than the extension of the spring at the end of step 1.

option C

Explanation: According to hooke's law, provided that the elastic limit is kept constant, the applied force (f) is proportional to extension (e).

Relative to our question, if a force is required to cause an extension (e) at the first instance, and at the second instance, the extension is 2 times of the initial, thus an additional force is required to make up for this increase in extension according to hooke's law.

From hooke's law,

f1/e1 = f2/e2

Where f1 = f, e1 = e, f2 =? and e2= 2e

f/e =f2/2e.

f ×2e = f2 ×e

f2 = f× 2e/ e

f2 = 2f.

As we can see the force in the second instance is twice the first.

Us = elastic potential energy = 1/2fe

Us1 = 1/2f1×e1

Us1 = 1/2× fe.

Us2 = 1/2 ×f2×e2

Us2 = 1/2 × 2f ×2e

Us2 = 4 ×fe/2

Us2 = 2 ×fe

If from Us1, we have that fe =2Us1 and we substitute this into Us2, we have that

Us2 = 2 (2Us1)

Us2 = 4Us1.

If Us2 is 4 times Us1, we can say that Us2 is greater than twice of Us1 which validates option c

0.11 m/s and 0.28 m/s should be correct