A block of mass 2kg is kept at origin at t=0 and is having velocity 4√5 ms- 1 in positive x-direction. The only force acting on it is a conservative and its potential energy is defined as U=-x3 + 6x2 + 15 (SI units). Its velocity when its acceleration is minimum after t=0 is

Question

A block of mass 2kg is kept at origin at t=0 and is having velocity 4√5 ms- 1 in positive x-direction. The only force acting on it is a conservative and its potential energy is defined as U=-x3 + 6x2 + 15 (SI units). Its velocity when its acceleration is minimum after t=0 is (A) 8 ms- 1 (B) 4 ms- 1 (C) 10√24 ms- 1 (D) 20 ms- 1

Tardigrade Admin · Accepted Answer

The expression of force can be calculated as

F = - \frac{d U}{d x} \Rightarrow F = 3 x^{2} - 12 x

Now, for minimum force,

\Rightarrow \frac{d F}{d x} = 0 \Rightarrow 6 x - 12 = 0 \Rightarrow x = 2 m

Therefore, the force as well as the acceleration is minimum at

x = 2

From energy conservation,

U_{i} + K_{i} = U_{f} + K_{f}

\Rightarrow [15] + [\frac{1}{2} \times 2 \times 80] = [- (2)^{3} + 6 \times (2)^{2} + 15] + [\frac{1}{2} \times 2 \times v^{2}] \Rightarrow 15 + 80 = 31 + v^{2}

\Rightarrow v^{2} = 64

\Rightarrow v = 8 m s^{- 1}

$8 m s^{- 1}$

$4 m s^{- 1}$

$1024 m s^{- 1}$

$20 m s^{- 1}$

Q. A block of mass 2kg is kept at origin at t=0 and is having velocity 45​ms−1 in positive x-direction. The only force acting on it is a conservative and its potential energy is defined as U=−x3+6x2+15 (SI units). Its velocity when its acceleration is minimum after t=0 is

8ms−1

4ms−1

1024​ms−1

20ms−1

$8 m s^{- 1}$

$4 m s^{- 1}$

$1024 m s^{- 1}$

$20 m s^{- 1}$