Question

A body of mass 20 kg is initially at a height of 3 m above the ground . It is lifted to a height of 2 m from that position. Its increase in potential energy is ​

Answer 1

Answer:

400

Explanation:

Potential energy=mgh

Here ,

initial m=20kg

initial h=3m

g=10 m/s

final h=5m

∴initial potential energy=20×3×10

=600

final potential energy=20×5×10

=1000

∴Increase in potential energy=1000-600

=400

Answer 2

Given:

A body of mass 20 kg is initially at a height of 3 m above the ground . It is lifted to a height of 2 m from that position.

To find:

Increase in potential energy (i.e. ∆PE) ?

Calculation:

• Initial height = 3 m

So, initial PE:

$PE1 = mgh = 20 \times 10 \times 3 = 600 \: J$

• New height = 3 + 2 = 5 m.

So, new PE:

$PE2 = mgh = 20 \times 10 \times 5= 1000 \: J$

So, change in PE :

$\Delta PE = PE2 - PE1$

$\implies\Delta PE = 1000 - 600$

$\implies\Delta PE = 400 \: J$

So, increase in PE is 400 Joule.