Question

Q1.A particle of mass 5 kg is at the top of a building of height 20 m.

a. What is the potential energy of the particle with respect to ground?

b. If the block is released from there, what will be its potential energy c. Calculate its velocity after one second using kinematical equations.

d. Now calculate its kinetic energy.

after one second!

e. Can you observe any relation between answers of part a, b and d? Explain​

Answer 1

Explanation:

a). Potential Energy w.r.t ground is:

PE = m×g×h

PE = 5 × 10 ×20

PE = 1000joule

b) When block is released, the velocity after 1 second will be :

∴v=u+at

$\implies\: v =0 + g(1)$

$\implies\: v =10 \times 1$

$\implies\: v =10 \: m {s}^{ - 1}$

d) So, kinetic energy will be :

$\therefore \: KE = \dfrac{1}{2} m {v}^{2}$

$\implies \: KE = \dfrac{1}{2} \times 5 \times {(10)}^{2}$

⟹KE=250joule

Yes, the relationship between a, b and d is:

• Sum of potential energy and kinetic energy of the object at any point with respect to the ground remain constant.

Answer 2

A particle of mass 5 kg is at the top of a building of height 20m.