The kinetic energy of a body is 2500 KJ corresponding to a velocity of 700 m/sec.

Estimate the loss in kinetic energy when its velocity is reduced to 450m/sec

Answer :

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Explanation:

To estimate the loss in kinetic energy when the velocity of a body is reduced from 700 m/s to 450 m/s, we can use the formula for kinetic energy:

\[ KE = \frac{1}{2}mv^2 \]

where:

- \( KE \) is the kinetic energy,

- \( m \) is the mass of the body,

- \( v \) is the velocity of the body.

Given:

- Initial velocity, \( v_1 = 700 \) m/s

- Initial kinetic energy, \( KE_1 = 2500 \) kJ (convert to joules: \( 1 \text{ kJ} = 1000 \text{ J} \), so \( KE_1 = 2500 \times 1000 = 2,500,000 \) J)

- Final velocity, \( v_2 = 450 \) m/s

- We need to find the final kinetic energy, \( KE_2 \).

1. Calculate the initial kinetic energy \( KE_1 \):

\[ KE_1 = \frac{1}{2}mv_1^2 \]

\[ 2,500,000 = \frac{1}{2}m \times (700)^2 \]

\[ 2,500,000 = \frac{1}{2} \times m \times 490,000 \]

\[ m \approx \frac{2,500,000}{245,000} \approx 10.2 \text{ kg} \]

2. Calculate the final kinetic energy \( KE_2 \):

\[ KE_2 = \frac{1}{2}mv_2^2 \]

\[ KE_2 = \frac{1}{2} \times 10.2 \times (450)^2 \]

\[ KE_2 = \frac{1}{2} \times 10.2 \times 202,500 \]

\[ KE_2 = \frac{1}{2} \times 2,060,500 \]

\[ KE_2 = 1,030,250 \text{ J} \]

3. Calculate the loss in kinetic energy:

\[ \text{Loss in } KE = KE_1 - KE_2 \]

\[ \text{Loss in } KE = 2,500,000 - 1,030,250 \]

\[ \text{Loss in } KE = 1,469,750 \text{ J} \]

Therefore, the loss in kinetic energy when the velocity of the body is reduced from 700 m/s to 450 m/s is approximately \( 1,469,750 \) joules.

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