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प्रश्न
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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उत्तर
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/s2
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/s2
F = ma = − 5 × 10−2 × 5
⇒ F = − 0.25 N along the motion
or, F = 0.25 N opposite the motion.
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