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# Tag Archives: Potential energy

## Buoyancy

A pound coin floating in mercury (Photo credit: Wikipedia)

I read in one of my chapters that when a body is immersed in a container of water (kept on a weighing machine which  reads ‘W’ at first )the apparent weight lost by the body is equal to the buoyant force, and thus the reading ‘W’ increases.

Then I came across a question like this –

A beaker containing water kept on a weighing machine weighs W. A body of weight  ‘w’ is dropped in it. It is floating & experiencing a buoyant force B, then the reading on machine is –

(a)w+W
(b)W+B
(C)W+w-B
(d)W

the book says answer is (a)
but I think it is (c)

(Anwesha posted this questions)

The total force acting downwards is W+w and the weighing scale is providing an equal reaction. the normal reaction offered by the weighing scale is what we get as the reading.

therefore the reading on the weighing scale must be W+w

The buoyant force is acting on the object dropped and is not contributing to the normal reaction offered by the weighing scale. ## Where is the light energy going?

where is the sun’s light enery going once the sun has set?

or

where is the light going when we switch off the torch facing an object. if light gets reflected how long it will do so. and if it gets absorbed, then will it add to the potential energy of the object that is facing the torch?

or Is there any cycle (like hydrocycle)to prove that energy can neither be created nor be destroyed? ## Bouncing ball and energy dissipation problem

A player throws a ball of mass 0.6 kg into the air, to a height of 4.0 m above the ground. The ball then falls to the ground. During the impact, 22% of the ball’s energy is lost. Calculate the height to which the ball rises after the bounce. Your assistance will be appreciated..

Mustafa posted

m= 0.6 kg

h = 4.0 m

Energy = mgh

During impact 22% energy is lost. Therefore, energy remaining = 88%

Therefore the energy with which the ball rebounds = 24 x 88/100 J

This is equal to the potential energy at the highest point after bouncing

If the new height is h’

mg h’ = 88% mgh

h’ = 0.88 x h = 0.88 x 4.0 =3.52 m ### Visitors So Far @ AskPhysics

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