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Question

A block of mass m kg is attached to a massless spring of spring constant k N/m. This system is accelerated upwards with an acceleration a m/s2. Then the elongation in spring will be (in metres):

A
mgk
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B
m(ga)k
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C
m(g+a)k
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D
mak
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Solution

The correct option is C m(g+a)k

From the FBD of block, applying Newton's 2nd law along direction of acceleration:
Tmg=ma......(1)
T=mg+ma ......(2)

Let the elongation in spring be x m, hence spring force kx will act in the form of tension T on the block.
kx=T ........(3)

From Eq.(2) and Eq.(3), we get:

kx=m(g+a)
x=m(g+a)k

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