Question

Please answer my questions fast....

Consider a drop of rain water having mass 1g falling from a height of 1 km. It hits the ground with a speed of 50m/s. Find the work done by the (i) gravitational force and the (ii) resistive force of air.​

Answer 1

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Answer 2

So now I can answer freely

Actually I was banned on brainly for 48 hours

Ok so now ,

Mass = 1g = 1/1000 kg

Falling from height = 1 km 1000 m

Work done = force * displacement

For this question you need to know work energy theorem given as:-

Worked done by all the forces either internal or external , either conservative or non conservative (I said all the forces) = change in total kinetic energy of the body

This theorem is universal (as universal as Newton's laws of motion)

( you will read this in 11th under the chapter work energy and power )

Also for this question without thinking about anything (I don't know from where you are asking such questions because what about boards? )

Just keep one thing in mind that work done by the gravity g on a particle of mass m= mgh

Because force = mg

And displacement caused is just change in vertical height = h

mgh = (1/1000 kg * 10 m/s^2 * 1000 m) = 10 J

W by gravity + W by resistive force (because total force acting on particle is only Gravitational force and air resistance) = Wg + Wr

Wg =W by gravity

Wr =W by restive force

By work energy theorem :-

Wg + Wr = change in total kinetic energy

Now kinetic energy is given by 1/2mV^2

mgh + Wr = 1/2 mV^2 - 1/2 mU^2

Where ,

V = final velocity = 50 m/s

U = initial velocity = 0 (because rain was dropping meaning started motion with zero velocity )

(1/1000 kg * 10 m/s^2 * 1000 m) + Wr = 1/2 (1/1000 kg)*(50 m/s) ^2

10 J + Wr = 5/4 J

Wr = - 35/4 J