The force F acting on a particle moving in a straight line as shown in Figure. What is the work done by the force on the particle in the first meter of the trajectory?

5 J

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10 J

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15 J

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2.5 J

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Solution

The correct option is D $2.5 J$
Shaded area = Work done by the force in first meter of trajectory
$= \frac{1}{2} \times 5 \times 1$
$= 2.5 J$

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The force F acting on a particle moving in a straight line as shown in Figure. What is the work done by the force on the particle in the first meter of the trajectory?

The correct option is D 2.5JShaded area = Work done by the force in first meter of trajectory=12×5×1=2.5J

The correct option is D $2.5 J$
Shaded area = Work done by the force in first meter of trajectory
$= \frac{1}{2} \times 5 \times 1$
$= 2.5 J$