AP EAMCET · PHYSICS · Work Power Energy
If a particle of mass ' m ' covers half of the horizontal circle with constant speed ' v ', then the change in its kinetic energy is
- A \(m v^2\)
- B Zero
- C \(2m v^2\)
- D \(\frac{1}{2} m v^2\)
Answer & Solution
Correct Answer
(B) Zero
Step-by-step Solution
Detailed explanation
\(\Delta KE = KE_{final} - KE_{initial}\) \(\Delta KE = \frac{1}{2} mv^2 - \frac{1}{2} mv^2\) \(\Delta KE = 0\)
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