Home/Question #4724
JEE AdvancedMedium

Q.A particle moves under a force F(x)=ax3F(x)=ax^3. If it starts from rest at x=0, its speed at position x is

a
ax42m\sqrt{\frac{ax^4}{2m}}
b
ax44m\sqrt{\frac{ax^4}{4m}}
c
ax4m\sqrt{\frac{ax^4}{m}}
d
ax2m\sqrt{\frac{ax^2}{m}}

Correct Answer: Option B

The correct solution involves applying the fundamental concept to derive the final value step by step...

Unlock Detailed Solution

Register for Free

Already a member? Login here