Question

A body of mass m which is moving with velocity has kinetic energy E.the same body moves with double the initial velocity then its kinetic energy becomes​

Answer 1

Answer:

K.E.=1/2×m×v^2

= 1/2 ×m × 2^2

It will become 4 times that of before

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Answer 2

Answer:

• The Kinetic Energy Becomes 4 times of the Initial one

Given:

Case-1

1. The Initial velocity be " u₁ "
2. Mass of the body " m "
3. Let the Kinetic Energy be " K.E₁ "

Case-2

1. Let the change in initial velocity be " u₂ "
2. Mass of the body " m "
3. Let the change in Kinetic Energy be " K.E₂ "

Explanation:

$\rule{300}{1.5}$

From the formula we know,

⇒ K.E = 1 / 2 m u²

Where,

• K.E Denotes Kinetic Energy.
• m Denotes Mass.
• u Denotes velocity.

Now,

⇒ K.E = 1 / 2 m u²

From above reference we get,

⇒ K.E ∝ u²

Therefore, it becomes,

⇒ K.E₁ / K.E₂ = u₁² / u₂²

From the statement of question we know,

Body moves with double the initial velocity

So,

⇒ u₂ = 2 u₁

Substituting in the above relation.

⇒ K.E₁ / K.E₂ = u₁² / (2 u₁)²

⇒ K.E₁ / K.E₂ = u₁² / 4 u₁²

⇒ K.E₁ / K.E₂ = 1 / 4

Rearranging,

⇒ K.E₂ = 4 × K.E₁

⇒ K.E₂ = 4 K.E₁

⇒ K.E₂ = 4 K.E₁

∴ The Kinetic Energy Becomes 4 times of the Initial one.

$\rule{300}{1.5}$