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A 3 kg object has initial velocity 6i^-2j^ m s-1. What will be the total work done (in joule) on the object if its velocity changes to (8i^+4j^) m s-1?
PhysicsWork Power EnergyJEE Main
Solution:
2108 Upvotes Verified Answer
The correct answer is: 60
The net work done on the object is equal to the change in the kinetic energy of the object

Wnet= Kf-Ki=ΔK

Kinetic energy K=12mv2

Velocity v2=vx2+vy2

=( 36+4 ) m 2 /s 2

v 2 =40 m 2 /s 2

Kf=12×3×40=60 J

Again, velocity v=82+42

v=64+16=80  m/s

Kinetic energy Kf=123×80  m 2 /s 2 =120 J

From work energy theorem

Wnet= Kf-Ki

=120 J-60 J=60 J

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