Question

The potential energy of a particle as a function of position will be given as

U(x) = A x2 + B x + C,

where U will be in joules when x is in meters. A, B, and C are constants.
What is the force on this particle, in newtons, at x = 38 cm, if the constants are A = 1.2 J/m2, B = 4.2 J/m, and C = 5.9 J?

Answer 1

To calculate the force on a particle at a specific position x using the potential energy function U(x), differentiate U(x) concerning x, then substitute the given position. The result is a force of -5.112 Newtons at x = 0.38 m.

Step-by-step explanation:

To find the force on a particle given the potential energy function U(x) = A x² + B x + C, we use the concept that the force exerted on a particle at a position x is the negative derivative of the potential energy concerning position, F(x) = -dU/dx.

First, we find the derivative of the given potential energy function:

• dU/dx = 2Ax + B

Then we plug in our constants and position (38 cm converted to meters, which is 0.38 m):

• F(x) at x = 0.38 m = -(2(1.2 J/m²)(0.38 m) + 4.2 J/m)

Calculating, we find:

• F(0.38) = -(2(1.2)(0.38) + 4.2)N = -(0.912 + 4.2)N = -5.112 N

Therefore, the force on the particle at x = 0.38 m is -5.112 Newtons.

Answer 2

The force on the particle at x = 38 cm with given constants A, B, and C is -3.432 newtons, which shows that the force acts in the direction opposite the increase in x.

Step-by-step explanation:

The force on a particle given a potential energy function of U(x) = Ax2 + Bx + C is found by taking the negative gradient of the potential energy with respect to position x. From the question, A = 1.2 J/m2, B = 4.2 J/m, and C = 5.9 J. The force is thus given as F(x) = -dU/dx = -2Ax - B. At x = 38 cm (0.38 m), the force is F(0.38) = -2(1.2)(0.38) - 4.2 which simplifies to F(0.38) = -3.432 N. The negative sign indicates that the force is directed opposite to the increase in the position along the x-axis.