Test Index

Simple Harmonic Motion

Section: Physics

© examsnet.com

Question : 6

Total: 23

A particle of mass 1‌kg is subjected to a force which depends on the position as

)‌kg‌ms−2 with k=1‌kg‌s−2. At time t=0, the particle's position

)m and its velocity

)ms−1. Let vx and vy denote the x and the y components of the particle's velocity, respectively. Ignore gravity. When z=0.5m, the value of (xvy−yvx) is m2s−1

[JEE Adv 2022 P2]

Your Answer:

Go to Question:

Prev Question

Next Question

Similar Free Exams: