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Grade 10Mechanics

A single force acts on a particle in rectilinear motion. A plot of velocity versus time for the particle is shown in Fig . 11-28. Find the sign (positive or negative ) of the work done by the force on the particle in each of the intervals AB, BC, CD, and DE.

Hrishant Goswami

11 Years agoGrade 10

1 Answer

Jitender Pal

11 Years ago

A single force acts on a particle in rectilinear motion. A plot of velocity versus time for the particle is shown in the below figure.

In the interval AB, change in kinetic energy of the particle is positive. So kinetic energy of the particle is increasing, therefore the sign of the work done will be positive.
In the interval BC, change in kinetic energy of the particle is zero since initial velocity (v_i) of the particle is equal to the final velocity (v_f) of the particle in the interval BC. Therefore in this region work done will be zero.
In the interval CD, change in kinetic energy of the particle is negative. So kinetic energy of the particle is decreasing, therefore the sign of the work done will be negative.
In the interval DE, change in kinetic energy of the particle is positive. So kinetic energy of the particle is increasing, therefore the sign of the work done will be positive.

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