The kinetic energy of a body changes from 12 J to 60 J due to the action of a force of 5N on an object of mass 4 kg. The work done by the force is:

24 J

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60 J

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48 J

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36 J

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Solution

The correct option is C
48 J

ΔK.E. = work done = 60 - 12 = 48

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Q. There are two mutual perpendicular forces acting on an object. One force acts in the direction of motion of the object and other is perpendicular to the motion. During this interval, the kinetic energy of the body changes from

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A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.

Compute the

(a) work done by the applied force in 10 s,

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and interpret your results.

Q. Prove that the change in kinetic energy of a body is equal to the work done by an external force acting on the body ?

Q. A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results.

The work-energy theorem states that the work done by a force on a body is equal to the change in kinetic energy of the body.

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The kinetic energy of a body changes from 12 J to 60 J due to the action of a force of 5N on an object of mass 4 kg. The work done by the force is:

The correct option is C 48 J ΔK.E. = work done = 60 - 12 = 48

The correct option is C
48 J

ΔK.E. = work done = 60 - 12 = 48