Question

Part A The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. 3.0 m, 5.0 m 4.0 m, 5.0 m 2.0 m, 3.0 m 1.0 m

Answer 1

To find where the force on the mass is zero, differentiate the potential energy function and solve the resulting quadratic equation to get the x-values.

Step-by-step explanation:

To find the location(s) where the force on the mass is zero, we need to take the derivative of the potential energy function U(x) with respect to x which gives us the force as a negative gradient of the potential energy (-dU/dx). The given potential energy is U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. The force function is then F(x) = -dU/dx = -6.0x2 + 30x - 36. To find the values of x where F(x) = 0, we set this expression equal to zero and solve the quadratic equation.

The solutions to the quadratic equation of the form ax2 + bx + c = 0 are given by x = (-b ± sqrt(b2 - 4ac)) / (2a), which in this case translates to x = (30 ± sqrt(302 - 4*6*36)) / (2*6).

After calculating, we get the two x-values where the force on the mass is zero, which you can check against the given options.

Answer 2

The location are
`x_1 = 2 \ and \ x_2 = 3`

Step-by-step explanation:

From the question we are told that

The potential energy is
`U(x) = (2.0 \ J/m^3) * x^3 - (15 \ J/m) * x^2 + (36 \ J/m) * x - 23 \ J`

The force on the mass can be mathematically evaluated as

`F = - (d U(x))/(d x ) = -( 6 x^2 - 30x +36)`

The negative sign shows that the force is moving in the opposite direction of the potential energy

`F = - 6 x^2 + 30x - 36`

At critical point

`(d U(x))/(dx) = 0`

So

`- 6 x^2 + 30x - 36 = 0`

`- x^2 + 5x - 6 = 0`

Using quadratic equation formula to solve this we have that

`x_1 = 2 \ and \ x_2 = 3`