Question

A single conservative force `F(x)` acts on a `1.0-kg` particle that moves along the x-axis. The potential energy `U(x)` is given by `U(x)=20+(x-2)^2` where x is in meters. At `x=5.0m`, the particle has a kinetic energy of `20J`.
The maximum and minimum values of x, respectively, are
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 1

Correct Answer - A
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`