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प्रश्न
A constant force of 2⋅5 N accelerates a stationary particle of mass 15 g through a displacement of 2⋅5 m. Find the work done and the average power delivered.
उत्तर
Given:
\[ = 2 . 5 \times 2 . 5 = 6 . 25 J\]
Acceleration of the particle is,
\[a = \frac{F}{m} = \frac{2 . 5}{0 . 015}\]
\[ = \frac{500}{3} \text{ m/ s} ^2\]
Applying the work-energy principle for finding the final velocity of the particle,
\[\frac{1}{2}m v^2 - 0 = 6 . 25\]
\[ \Rightarrow \nu = \sqrt{\frac{6 . 25 \times 2}{0 . 15}} = 28 . 86 \text{ m/s }\]
So, time taken by the particle to cover 2.5 m distance,
\[t = \frac{\nu - u}{\alpha} = \frac{\left( 28 . 86 \right) \times 3}{500}\]
\[ \therefore \text{ Average power } = \frac{W}{t}\]
\[ = \frac{6 . 25 \times 500}{\left( 28 . 86 \right) \times 3} = 36 . 1 W\]
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