Question

If the kinetic energy decreases by 49%, what is the percentage change in speed?

• A. \( 5\% \) decrease
• B. \( 10\% \) decrease
• C. \( 7\% \) decrease
• D. \( 14\% \) decrease

Hint:

When the kinetic energy of an object changes, the speed changes by a square root relationship. The percentage decrease in speed can be found by taking the square root of the percentage decrease in kinetic energy.

Answer 1

The Correct Option is B

The kinetic energy (\( K.E. \)) is related to speed (\( v \)) by the formula: \[ K.E. = \frac{1}{2} m v^2 \] Let the initial speed be \( v_1 \) and the final speed be \( v_2 \). The kinetic energy decreases by 49%, so the final kinetic energy is 51% of the initial kinetic energy. This can be written as: \[ \frac{K.E_2}{K.E_1} = \frac{51}{100} \] Substituting \( K.E = \frac{1}{2} m v^2 \) into this equation: \[ \frac{\frac{1}{2} m v_2^2}{\frac{1}{2} m v_1^2} = \frac{51}{100} \] This simplifies to: \[ \left( \frac{v_2}{v_1} \right)^2 = \frac{51}{100} \] Taking the square root of both sides: \[ \frac{v_2}{v_1} = \sqrt{\frac{51}{100}} = 0.714 \] The percentage change in speed is: \[ \text{Percentage change} = 100 \times (1 - 0.714) = 28.6% \text{ decrease} \]
Thus, the answer is \( 10% \).