A block of mass mkg is attached to a massless spring of spring constant kN/m. This system is accelerated upwards with an acceleration am/s2. Then the elongation in spring will be (in metres):
A
mgk
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
m(g−a)k
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
m(g+a)k
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
mak
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Cm(g+a)k
From the FBD of block, applying Newton's 2nd law along direction of acceleration: T−mg=ma......(1) T=mg+ma ......(2)
Let the elongation in spring be xm, hence spring force kx will act in the form of tension T on the block. ⇒kx=T ........(3)