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Question

The figure shows the force (F) versus displacement (s) graph for a particle of mass m=2kg initially at rest:
(a) The maximum speed of the particle occurs at x=......m.
(b) The maximum speed of the particle occurs is ........ms−1.
(c) The particle once again has its speed zero at x=...........m.
284695_81b6de0b1e1048ca8a2486b78cc68a28.png

A
4.1833, 3, 6
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B
3, 4, 6
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C
6, 5, 3
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D
4, 5, 3
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Solution

The correct option is A 4.1833, 3, 6
The velocity of the particle will keep on increasing until it starts retarding
So, the maximum velocity of the particle will occur at x=3 m
As after that it starts deaccelerating
Now work done=x0Fdx=areaundertheforcedisplacementgraphtillx=3m
12×(5+10)+12×2×10=10+7.5=17.5
Let the maximum speed attained by the particle at x=3 m is 'v'
Accordingtoworkenergytheorem,
Work done on the system=change in the kinetic energy of the system=12×2×v2=17.5v2=17.5
v=17.5
Area of the graph under force displacement curve from x=3mtox=6mis15+202=17.5
So.workdonebytheforcebetweenthatintervalis17.5So,totalworkdonebytheforcetillx=6mis17.517.5=0So,kineticenergyoftheparticleatx=6mwillbezeroSo,velocityoftheparticlewillbezeroatx=6m

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