Question

A body of mass 1 Kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest after attaining a height of 18m. How much energy is lost due to air friction ? (g=10m/s^2)


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Answer 1

We have been given that,

Mass of the body = 1 Kg

Initial velocity = 20m/s

Maximum height = 18m

We have been asked to find the energy lost by the body due to air friction.

When the body is projected, it would be having Kinetic Energy.

When the body momentarily comes to rest, it would be having potential energy.

Kinetic Energy of the body would be:-

= $\dfrac{1}{2} mv^{2}$

= $\dfrac{1 \times 20 \times 20}{2}$

= 200 Joules

Potential Energy of the body would be:-

= mgh

= 1 x 10 x 18

= 180 Joules

All of the Kinetic Energy should have converted to Potential Energy but that didn't happen this means some of it was lost due to Air Friction.

Let the energy lost to air friction to be 'x' Joules.

According to question:-

KE = PE + x

200 = 180 + x

x = 200 - 180

x = 20 Joules

So, the energy lost due to air friction is 20 Joules !

Answer 2

Explanation:

Let us have a concept of Energy for body either in static or motion..

Potential Energy : Energy in its steady state (tends to stay in rest)

Potential energy = mgh

where m = mass , g = gravity due to acceleration , h = height above surface level.

Kinetic Energy : Energy in its motion..(tends to stay in motion)

Kinetic energy = ½mv²

where m = mass , v = velocity.

Given

mass = 1 kg

velocity = 20 m/s

height = 18 m

gravity = 10 m/s²

since when thrown ,

Kinetic energy = ½ mv² = ½×1×20² = 200 J

Potential energy = mgh = 1×10×18 = 180 J

So energy loss = KE - PE = 200-180 = 20 J