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`.
Determine the equation of `F(x)` as a function of x.

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 . What is the mechanical energy of a system?

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 kinetic energy of the particle and the value of x at which maximum kinetic energy occurs are

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 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 ?.

A single conservative f_((x)) acts on a m = 1 kg particle moving along the x-axis. The potential energy U_((x)) is given by : U_((x)) = 20 + (x - 2)^(2) where x is in metres. At x = 5 m , a particle has kintetic energy of 20 J The maximum speed of the particle is:

A single conservative f_((x)) acts on a m = 1 kg particle moving along the x-axis. The potential energy U_((x)) is given by : U_((x)) = 20 + (x - 2)^(2) where x is in metres. At x = 5 m , a particle has kintetic energy of 20 J The minimum speed of the particle is:

A single conservative f_((x)) acts on a m = 1 kg particle moving along the x-axis. The potential energy U_((x)) is given by : U_((x)) = 20 + (x - 2)^(2) where x is in metres. At x = 5 m , a particle has kintetic energy of 20 J The largest value of x will be:

A single conservative f_((x)) acts on a m = 1 kg particle moving along the x-axis. The potential energy U_((x)) is given by : U_((x)) = 20 + (x - 2)^(2) where x is in metres. At x = 5 m , a particle has kintetic energy of 20 J The position of equilibrium and its nature is:

A single conservative f_((x)) acts on a m = 1 kg particle moving along the x-axis. The potential energy U_((x)) is given by : U_((x)) = 20 + (x - 2)^(2) where x is in metres. At x = 5 m , a particle has kintetic energy of 20 J The maximum value of potential energy is:

A single conservative f_((x)) acts on a m = 1 kg particle moving along the x-axis. The potential energy U_((x)) is given by : U_((x)) = 20 + (x - 2)^(2) where x is in metres. At x = 5 m , a particle has kintetic energy of 20 J The least value of x (position of particle is ) will be: