Question

A body of mass 500 g is taken from a height of 2.5 m to 3 m. Find the increase in potential energy of the body. Take g = 10 N kg-1 .

Answer 1

Answer :

Mass of body = 500g = 0.5kg

Initial height = 2.5m

Final height = 3m

We have to find the increase in potential energy of the body$.$

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◈ Potential energy of body of mass m fat height H from ground point is given by

• U = mgH

It is a scalar quantity having only magnitude. Potential energy is measured wrt to reference point and at reference point it is taken zero.

➝ ΔU = U₂ - U₁

➝ ΔU = mg(H₂ - H₁)

➝ ΔU = 0.5 × 10(3 - 0.5)

➝ ΔU = 5(2.5)

➝ ΔU = 12.5J

Hence, increase in potential energy of body is 12.5J.

Answer 2

Question:-

A body of mass 500 g is taken from a height of 2.5 m to 3 m. Find the increase in potential energy of the body.

{ Take g = 10 m/s² }

Given:-

A body of mass 500 g is taken from a height of 2.5 m to 3 m. {Take g = 10 m/s²}

Find:-

Increase in potential energy of the body.

Solution:-

Gravitational potential energy of body is denoted by

${ \sf{{ \purple{U} }\: \: or \: \: { \purple{P.E_{G} }}= mgh}}$

where,

=> m = mass of body

=> g = acceleration due to gravity

=> h = heigth above the surface

Hence,

=> m = 500 g = 0.5 kg ...... ( 1 )

{ According to question }

=> g = 10 m/s² ...... ( 2 )

Height of body change from 2.5 m to 3 m .

Let, initial height = h1 = 2.5 m and

final height = h2 = 3 m

change in height = Δh = h2 - h1

=> Δh = 3 - 2.5

=> Δh = 0.5 m ..... ( 3 )

Now we use formula to find the increase in potential energy of the body.

${ \sf{ \red{\dashrightarrow}{ \purple{P.E_{G} }}= mgΔh}}$

{ from ( 1 ) , ( 2 ) & ( 3 ) }

${ \sf{ \red{\dashrightarrow}{ \purple{P.E_{G} }}=0.5 \times 10 \times 0.5 }}$

${ \sf{ \red{\dashrightarrow}{ \purple{P.E_{G} }}= 2.5 \: Joule}}</p><p>$

Hence increase in potential energy of the body is 2.5 Joule.