Question
A constant force of 2.50N accelerates a stationary particle of mass 15g through a displacement of 2.50m. Find the work done and the average power delivered.
✓
Answer
F = 2.50N, S = 2.5m, m = 15g = 0.015kg.So, $\text{w} =\text{F} \times \text{S}$
$\Rightarrow\text{a}=\frac{\text{F}}{\text{m}}=\frac{2.5}{0.015}$
$=\frac{500}{3}\text{m/s}^2$
$=\text{F}\times\text{S}\cos0^\circ$ (acting along the same line)
$=2.5\times2.5=6.25\text{J}$
Let the velocity of the body at b = U. Applying work-energy principle $\frac{1}{2}\text{mv}^2-0=6.25$
$\Rightarrow\text{V}=\sqrt{\frac{6.25\times2}{0.015}}=28.86\text{m/sec}$
So, time taken to travel from A to B.
$\Rightarrow\text{t}=\frac{\text{v-u}}{\text{a}}=\frac{28.86\times3}{500}$
$\therefore$ Average power $=\frac{\text{W}}{\text{t}}=\frac{6.25\times500}{(28.86)\times3}=36.1$
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