Home/Question #4674
JEE AdvancedMedium

Q.A particle moves along x-axis under force F(x)=axbx2F(x)=ax-bx^2. The work done from x=0 to x=a/b is

a
a36b2\frac{a^3}{6b^2}
b
a33b2\frac{a^3}{3b^2}
c
a32b2\frac{a^3}{2b^2}
d
a3b2\frac{a^3}{b^2}

Correct Answer: Option A

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