If the work done in increasing the velocity of a body by 25% is 900 J, then the work done in decreasing the velocity of the same body by 25% is

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q88 – AP-EAMCET 2026

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Home Physics Work, Energy & PowerQuestion #642

If the work done in increasing the velocity of a body by

25 %

900 J

, then the work done in decreasing the velocity of the same body by

25 %

Clue 1 of 3: Work-Energy Theorem

The work done on a body is equal to the change in its kinetic energy.

🧠 Full Solution Path

Step 1: Define Initial and Final Kinetic Energies✦ Active

Let the initial velocity of the body be $v$ and its mass be $m$ . The initial kinetic energy is $K E_{i} = \frac{1}{2} m v^{2}$ .

When the velocity increases by $25 %$ , the final velocity $v_{1}$ is $v + 0.25 v = 1.25 v = \frac{5}{4} v$ . The final kinetic energy $K E_{1}$ is $\frac{1}{2} m (v_{1})^{2} = \frac{1}{2} m (\frac{5}{4} v)^{2} = \frac{25}{16} (\frac{1}{2} m v^{2}) = \frac{25}{16} K E_{i}$ .

When the velocity decreases by $25 %$ , the final velocity $v_{2}$ is $v - 0.25 v = 0.75 v = \frac{3}{4} v$ . The final kinetic energy $K E_{2}$ is $\frac{1}{2} m (v_{2})^{2} = \frac{1}{2} m (\frac{3}{4} v)^{2} = \frac{9}{16} (\frac{1}{2} m v^{2}) = \frac{9}{16} K E_{i}$ .

💡 Teacher's Secret Hint

Remember to square the entire final velocity term, including the fractional or decimal multiplier, when calculating the final kinetic energy.

Step 2: Calculate Initial Kinetic Energy from Given Work Done○ Expand

The work done in increasing the velocity by $25 %$ is $W_{1} = K E_{1} - K E_{i}$ . We are given $W_{1} = 900 J$ .

W_{1} = \frac{25}{16} K E_{i} - K E_{i} = (\frac{25}{16} - 1) K E_{i} = \frac{9}{16} K E_{i}

Substitute the given value of $W_{1}$ :

900 J = \frac{9}{16} K E_{i}

Solve for $K E_{i}$ :

K E_{i} = 900 \times \frac{16}{9} = 100 \times 16 = 1600 J

💡 Teacher's Secret Hint

Ensure you correctly handle the fraction when isolating $K E_{i}$ . Multiplying by the reciprocal is a common way to do this.

Step 3: Calculate Work Done for Decreasing Velocity○ Expand

The work done in decreasing the velocity by $25 %$ is $W_{2} = K E_{2} - K E_{i}$ .

W_{2} = \frac{9}{16} K E_{i} - K E_{i} = (\frac{9}{16} - 1) K E_{i} = - \frac{7}{16} K E_{i}

Substitute the value of $K E_{i} = 1600 J$ :

W_{2} = - \frac{7}{16} \times 1600 J = - 7 \times 100 J = - 700 J

The negative sign indicates that work is done *by* the body (or an external agent does negative work) to decrease its kinetic energy. The magnitude of the work done is $700 J$ .

💡 Teacher's Secret Hint

Work done can be negative, indicating that the net force on the object is opposite to its displacement, or that the object is doing work on its surroundings, leading to a decrease in its kinetic energy. In multiple-choice questions, if only positive options are given, the magnitude of work is usually expected.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 88 – AP-EAMCET 2026

Clue 1 of 3: Work-Energy Theorem

🧠 Full Solution Path