Question

A body of mass m is thrown at an angle α to the horizontal with the initial velocity v0. Find the mean power developed by gravity over the whole time of motion of the body, and the instantaneous power of gravity as a function of time.

Answer 1

Solution

The velocity of body after t sec is given by

$\overrightarrow{ v}=\overrightarrow{ v_0}+ \overrightarrow{g} t$

Power is given by

$P=\overrightarrow{F}. \overrightarrow{v}=m\overrightarrow{g}. \overrightarrow{v}$

Here (.) denotes dot product.

$P= m \overrightarrow{g} .( \overrightarrow{ v_0}+ \overrightarrow{g} t)$

$P=mg(gt-v_0 \sin \alpha)$

Then mean power

$<P>=\frac{E}{\tau}=\frac{m \overrightarrow{g} .\Delta \overrightarrow{ r}}{\tau}$

Because mg and $\Delta r$ are perpendicular to each other so

Mean power

<P>=0