Sum

A particle of mass 50 g moves in a straight line. The variation of speed with time is shown in the following figure. Find the force acting on the particle at t = 2, 4 and 6 seconds.

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#### Solution

Given:

Mass of the particle, m = 50 g = 5 × 10^{−2} kg

Slope of the v-t graph gives acceleration.

At t = 2 s,

Slope = \[\frac{15}{3} = 5 m/ s^2\]

So, acceleration, a = 5 m/s^{2}

F = ma = 5 × 10^{−2} × 5

⇒ F = 0.25 N along the motion.

At t = 4 s,

Slope = 0

So, acceleration, a = 0

⇒ F = 0

At t = 6 sec,

Slope =\[\frac{- 15}{3} = - 5 m/ s^2\]

So, acceleration, a = − 5 m/s^{2}

F = ma = − 5 × 10^{−2} × 5

⇒ F = − 0.25 N along the motion

or, F = 0.25 N opposite the motion.

Concept: Newton’s Second Law of Motion

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