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Question

A 2 kg body (initially at rest) is displaced by a variable force F=(3x2+15) N on a rough horizontal plane with coefficient of kinetic friction μ=0.5. Find the speed of the block, after the block has travelled a distance of 5 m.

A
v=53 m/s
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B
v=3 m/s
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C
v=5 m/s
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D
v=56 m/s
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Solution

The correct option is D v=56 m/s
Mass of body, m=2 kg
Force acting on the body F=(3x2+15) N
Coefficient of kinetic friction, μ=0.5
Due to sliding on the horizontal surface, kinetic friction will be acting on the body in a direction opposite to its motion.
fk=μN=μmg

Work done by friction on the body,
Wf=Fdcosθ=(μmg)×x×(cos180)
Wf=μmgx
Wf=(0.5)×2×10×5=50 J

Work done by variable force (F) on the body is
WF=Fdx
Putting the limits from x=0x=5 m,
WF=x=5x=0(3x2+15)dx
=[3×x33]50+[15x]50
=(1250)+(750)=200 J

Work done by gravity Wmg=0 (because displacement is horizontal)

Thus, total work done by the external forces on the body is:
Wext=WN+Wmg+Wf+WF
Wext=0+050+200=150 J ...(i)
Initial kinetic energy, KEi=0
Let v be the velocity at x=5 m.
Final kinetic energy KEf=12mv2=12×2×v2
ΔKE=KEfKEi=v2 ....(ii)

Applying work energy theorem on the body:
Wext=ΔKE
From Eq (i)& (ii),
150=v2
v=150=56 m/s

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