wiz-icon
MyQuestionIcon
MyQuestionIcon
4
You visited us 4 times! Enjoying our articles? Unlock Full Access!
Question

Let a pring with force constant k is compressed by x1. Stored potential energy of the spring U1 = 12 k x2 . Let it be further compressed by x2. Let it be further compressed by x2 . Change in potential energy of the spring


A

k ( - )

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

k [ (() - )]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

k

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

k

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B

k [ (() - )]


ΔU = Uf - Ui

Uf=12K(x1+x2)2

Ui = 12K x12

Δ U= 12 K (x1+x2)2 - 12 k x12


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Springs: Playing with a Slinky
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon