Question

A body of mass 5kg is taken from a height of 3m to 6m then what is the increase in the potential energy of a body

Answer 1

$\maltese \: \red{\mid{\underline{\overline{\textbf{Answer}}}\mid}}$

Potential energy of a body is the energy possessed by a body due to it's position. It is defined as the product of mass height of object from ground and the acceleration due to gravity. Represented by the equation.

$\maltese \: \orange{\mid{\underline{\overline{\textbf{pe = m × g × h}}}\mid}}$

Now, it is given that :

• Mass of body = 5kg
• Initial height = 3m
• Final height = 6m.
• ∆Potential Energy = ?

Now the change in potential energy will be potential energy at final height - potential energy at initial height.

The potential energy of body at initial height :

$\maltese \: pe = 5 \times 9.8 \times 3$

$\maltese \: pe = 49 \times 3$

$\maltese \: pe = 147j$

Now potential energy of object at final height :

$\maltese \: pe = 5 \times 9.8 \times 6$

$\maltese \: pe = 49 \times 6$

$\maltese \: pe = 294j$

Now the change in potential energy in this process will be :

∆PE = 294J - 147J

∆PE = 147J

Therefore the potential energy of body increases by 147J , or we can also say that the potential energy gets doubled.

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Answer 2

Change in potential energy

= mgh′ - mgh

{h′ = final height; f = initial height},

= mg(h′ - h)

We have,

Mass of object = 5 kg

Initial height = 3 m

Final height = 6 m

g = 9.81 m/s²

∴ Change = (5 kg)(9.81 m/s²)(6 m - 3 m)

⇒ PE_change = (5 kg)(9.81 m/s²)(3 m)

⇒ PE_change = 147.15 Nm

⇒ PE_change = 147.15 J {Answer}