A single conservative force F(x) acts on a on a (1.0kg ) particle that moves along the x-axis. The potential energy U(x) is given by:
`U(x) =20 + (x-2)^(2)`
where, x is meters. At `x=5.0m` the particle has a kinetic energy of `20J`
(a) What is the mechanical energy of the system?
(b) Make a plot of U(x) as a function of x for `-10mlexle10m`, and on the same graph draw the line that represents the mechanical energy of the system. Use part (b) to determine.
(c) The least value of x and
(d) the greatest value of x between which the particle can move.
(e) The maximum kinetic energy of the particle and
(f) the value of x at which it occurs.
(g) Datermine the equation for `F(x)` as a function of x.
(h) For what (finite ) value of x does `F(x) =0`?.

Options

A. (a) `7.38m, -3.38m`
B. (b) `6.38m, -4.38m`
C. (c) `7.38m, -2.83m`
D. (d) `6.38m, -2.38m`

Answer

A

Explanation

The greatest or least value of x will be at the point where K is zero and U is maximum.
`U_(max)=49Jimplies20+(x-2)^2=49implies(x-2)^2=29`
`impliesx-2=+-sqrt(29)impliesx=-3.38m, 7.38m`