A single conservative force Fx = (6.0x  12) N (x is in m) acts on a particle moving along the x axis. The potential energy associated with this force is assigned a value of +20 J at x = 0. What is the potential energy at x = 3.0 m?

Respuesta :

Answer:

[tex]U(3)=-43J[/tex]

Explanation:

Potential energy is minus the integral of Fdx.  Doing the integration yields:

[tex]U=\int\limits {6.0x+12}\, dx[/tex]

[tex]U=-3x^2-12x+C[/tex]

[tex]U(0)=20J[/tex]

so

[tex]U(0)=-3(0)^2+12(0)+C=20[/tex]

[tex]C=20[/tex]

Now for x=3.0m

[tex]U(3)=-3*(3)^2-12(3)+20[/tex]

[tex]U(3)=-43J[/tex]