Question

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.

 

Answer 1

Given:

\[\text{ F = 2 . 50 N, S = 2 . 5 m and m = 15 g = 0 . 015 kg} \]

Work done by the force,

 \[W = F \cdot S \cos 0^\circ\left( \text{ acting along the same line } \right)\]

\[ = 2 . 5 \times 2 . 5 = 6 . 25 J\]

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,

\[t = \frac{\nu - u}{\alpha} = \frac{\left( 28 . 86 \right) \times 3}{500}\]
\[ \therefore \text{ Average power } = \frac{W}{t}\]
\[ = \frac{6 . 25 \times 500}{\left( 28 . 86 \right) \times 3} = 36 . 1 W\]