Question

Consider a raindrop of mass 1.00 g falling from a height 1 km. It hits the ground with a speed of 50 m/s. Find work done by the opposing resistive force and also work done by the gravitational force.
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Answer 1

SOLUTION:

We will assume that the pebble is initially at rest on the cliff

So,

$\boxed{\red{u=0}}$

$\boxed{\purple{m=1g} = \red{10^{-3}kg}}}$

$\boxed{\red{v=50\ ms^{-1}}}$

$\boxed{\red{h=1km = 10^{3}m}}$

Now,

The change in KE of the pebble is

$\triangle K = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$

$\triangle K = \frac{1}{2}\times (10)^{-3} \times (50)^{2}- \frac{1}{2} \times(10)^{-3} \times (0)^{2}$

$\triangle K = \frac{1}{2}\times (10)^{-3} \times (50)^{2}- 0$

$\boxed{\triangle K = 1.25\ J }$

Now,

Assuming the g = 10 m/s² is constant, the work done by the gravitational force is

$W_{g} = F.h = mgh = 10^{-3} \times 10\times10^{3} = 10\ J$

Now,

If ($W_{r}$) is the work done by the resistive force on the pebble, then from the work-energy theorem,

$\triangle K = W_{g} +W_{r}$

OR

$W_{r}= \triangle K - W_{g} \\ W_{r} = 1.25 - 10\\ W_{r} = -8.7 J$

Hence,

• The work done by the unknown resistive force is -8.75 J