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A single conservative fore F(x) acts on ...

A single conservative fore `F_(x)` acts on a `2kg` particle that moves along the x-axis. The potential energy is given by.
`U =(x-4)^(2)-16`
Here, x is in metre and U in joule. At `x =6.0 m` kinetic energy of particle is `8 J` find .
(a) total mechanical energy
(b) maximum kinetic energy
(c) values of x between which particle moves
(d) the equation of `F_(x)` as a funcation is zero
(e) the value of x at which `F_(x)` is zero .

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
(a) At `x=6m`,
`U=(6-4)^(2)-16=-12 J`
`K=8 J`
`:. E=U + K =-4 J`
(b) `U_("min") =-16 J` at `x =4m`
`:. K_("max") =E-U_("min")`
`=-4 + 16`
`=12 J`
(c ) `K =0`
`:. U=E`
or `(x-4)^(2)-16 =-4`
or `x =(4+-2sqrt3) m`
(d) `F_(x) = -(dU)/(dx)=(8-2x)`
(e) `F_(x) =0 at x=4 m`.
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