Question

Potential Energy and Force

A mass m=3.00kg moving along the x-axis is acted on only by a single conservative force. The force has a potential energy function given by U(x)=(1.00J/m3)x3−(9.00J/m2)x2+(15.0J/m)x. It will be useful to graph this function on your calculator or computer.

Part A

Find the force on the mass as a function of x. (Leave the units out of the coefficients in your expression, and make sure all coefficients have three significant figures).

F(x)   =		N

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Part B

For this force, there are two equilibrium positions. Solve for them, and list them in increasing order below.

xeq   =		m

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Part C

Assume you are told that one of the turning points of the mass is at x=6.00m. Solve for the total energy of the system. (Recall that a turning point is defined as a position at which the object stops and turns around).  

E   =		J

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Part D

Determine if the mass will ever reach the first equilibrium point.

Determine if the mass will ever reach the first equilibrium point.

	There is not enough information to decide.
	The mass will reach the first equilibrium point.
	The mass will not reach the first equilibrium point.

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Part E

Solve for the speed of the mass when it is at the second equilibrium position.

veq,2   =		m/s

Answer 1

The potential energy

$F(x)=-\frac{dU(x)}{dx}=3x^2-18 x+15$

At equilibrium position

$F(x)=-\frac{dU(x)}{dx}=3x^2-18 x+15=0$

$\implies 3x^2-15 x-3x+15=0 \\ \implies (3x-15)x-(3x-15)=(x-1)(3x-15)=0$

Clearly two equlibrium points are

a turning point is defined as a position at which the object stops and turns around

velocity of the object as well as the kinetic energy is zero at turning point

Therefore the total energy is the potential energy at that turning point and that is given by

$E=U(x=6m)=6^3-9\times 6^2+15\times 6J=-18J$

$\frac{d^2U(x)}{dx^2}=6x-18$

at $x=1$ the second derivative is negative, means this is a maxima

at $x=5$ the second derivative is positive, means this is a minima

For stable equilibrium the body will tend to be in a minima.

at

at $x=5, U=125-9 \times 25+15 \times 5J=-25J$

For total energy $E=-18J$ , if the potential energy is $U=7J$   then the kinetic energy has to be   which is not possible as negative kinetic energy is absurd.

So The mass will not reach the first equilibrium point.

This potential energy will be converted to kinetic energy at equilibrium point

Therefore at second equilibrium position

For total energy $E=-18J$ , if the potential energy is $U=-25J$   then the kinetic energy has to be which is acceptable

$\frac{1}{2}mv^2=7J \\ \\ \implies v_{eq,2}=\sqrt{\frac{7 \times 2}{3}}m/s=2.16m/s$