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Question

A block of mass 100 kg is attached to the back of a truck with the help of a spring. The frictional coefficient between the truck and the block is 0.2 (take μs=μk). Acceleration of the truck is 5 m/s2 and the spring elongates by a distance of 1 cm. If the block is just about to slip, what is the force constant of the spring? Take g=10 m/s2.

A
300 N/m
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B
30000 N/m
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C
3000 N/m
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D
1000 N/m
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Solution

The correct option is B 30000 N/m
As the truck accelerates, spring will start to elongate so that it can provide the required force for acceleration in the xdirection.

Also, friction will act at its limiting value fmax to support the spring force, in order to avoid slipping between block and truck.

Solving from accelerated frame will ease the problem, since block will be at rest in that frame, but we will apply a Fpseudo in opposite direction on block in the FBD shown below:


Fpseudo=ma=100×5=500 N

Spring force =kx, where x=0.01 m= elongation in the spring

From the FBD, applying equilibrium condition in horizontal and vertical directions:

kx+fmaxFpseudo=0.......i
N=mg..............ii
and fmax=μsN=μsmg=200 N.....iii

Solving Eq. i , ii , iii gives:

k×(0.01)+200500=0

k=30000 N/m

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