Question

A block of mass 2 kg is kept at origin at t = 0 and is having velocity $4 \sqrt { 5 }$ m/s in positive x-direction. The only force acting on it is a conservative and its potential energy is defind as U = – x³ + 6x² + 15 (SI units). Its velocity when its acceleration is minimum after t = 0 is-

• A. 8 m/s
• B. 4 m/s
• C. 10$\sqrt { 24 }$m/s
• D. 20 m/s

Answer 1

Answer: (A)

8 m/s

Answer 2

F = – Ž F = 3x² – 12x

Now F = min. Ž = 0 Ž x = 2 m

U_i + K_i = U_f + K_f

15 + $\frac { 1 } { 2 }$ × 2 × 80 = [– (2)³ + 6 × (2)² +15] + $\frac { 1 } { 2 }$ × 2 × v²

v² = 64

v = 8m/sec