The problem for AP physics C students about the motion of the body by influence of a conservative force. If you can solve this problem, I would like to tutor you and give you a discount.

Question

A small object with mass m = 0.09 kg moves along the x-axis. The only force on the object is a conservative force that has potential energy function U(x) = - αx² + βx³ , where α = 2.00 J / m² and β = 0.300 J / m³.

The object is released from rest at small x. When the object is at x = 4.00 m, what are its

(a) speed,

(b) acceleration (magnitude and direction),

(c) what is the maximum value of x reached by the object during its motion?

Answer 1

We know how potential energy depends on x:

                                            U(x) = - αx² + βx³

We can find the force using the formula F(x) = - dU /dx, hence

                                   F(x) = - (-2αx + 3βx²)        or            F = 2αx - 3βx²

From another hand, according to Newton’s second law any force

                                              F = ma

From those 2 equations

                                             a(x) = (2αx - 3βx²) / m

Putting here all known values we get a = + 17.8 m/s². The plus sign tells us that vector of acceleration directed in the direction of x-axis.

To find speed, I will use the chain rule.

We know that a(x) = dv /dt. But we know only how acceleration a depends on x, but not t.

It means now I can apply the chain rule and write

                                                   a = dv/dx ∙dx/dt

Because v = dx/dt, I can write

                                                   a = (dv/dx)∙v

To find v from here, let’s separate variables and write

                                                vdv = a(x)dx 

Substituting for a, we have             

                                               vdv = (2αx - 3βx²)dx

Now we can use integration. I will integrate the left part for v changing from 0 to v and the right part for x from 0 to x. Then

                                              ∫ vdv = ∫(2αx - 3βx²)dx

Solving it, we are getting

                                                   v²/ 2 = αx² - βx³  

Hence, velocity

                                             v = √ 2(αx² – βx³)

At x = 4.00 m

                                             v = √ 2(2.00∙(4.00)² – 0.300∙(4.00)³

Hope for AP physics C student it is not hard to find the answer.

The body will stop moving, when its instantaneous speed is 0. So,

                                                 αx² – βx³ = 0

This equation is easy to solve

                                                x² ( α – βx) = 0

We are not interested in case when x = 0, instead we take second solution

                                                  x = α / β

Plugging in the values of coefficients, we get

                                         x = 2.00 J/m² / 0.300 J / m³ = 6.67 m.    

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