One end of a light spring of spring constnat k is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface.
In a displacement, the work done by the spring is $\frac{1}{2} {kx}^{2} .$ The possible cases are
a) the spring was initially compressed by a distance x and was finally in its natural length
b) it was initially streched by a distance x and finally was in its natural length
c) it was initially in its natural length and finally in a compressed position
d) it was initially in its natural length and finally in a stretched position.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

a, b are correct

b, c are correct

a, d are correct

b, d are correct

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Spring's stored elastic potential energy is equal to $\frac{1}{2} {kx}^{2}$ , where x represents the spring's compression or extension from its normal length. In this configuration, the spring can exert force equal to $kx$ on the block that is attached to it.

Hence the correct answer is a,b are correct.

Watch 3-min video & get full concept clarity