Question

The potential energy of 1kg particle free to move along the x-axis is given by
v(x) = [x⁴/4 -x²/2] j .
The total mechanical energy of the particle is 2j . then ,the maximum speed (in m/s) is .​

Answer 1

Explanation:- Given v(x) = (x⁴/4 -x²/2) j

For minimum value of v = dv/dx = 0

For minimum value of v = dv/dx = 0

• ⟹ 4x³/4 - 2x/2 = 0
• ⟹ x = 0 , x = +- 1

So, Vmin (x = +-1) = 1/4 - 1/2:= -1/4J

Now , Kmax + V min = total Machanical energy (2j)

• ⟹ Kmax = (1/4) +2
• ⟹ Kmax = 9/4
• ⟹ mv²/2 = 9/4
• ⟹ v² = 18/4 [Given m = 1 kg ]
• ⟹ v =√9/4 = 3/√2 Answer

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