Calculating Potential Energy for 16 kg Object

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The potential energy of a 16 kg object at x = 2.0 m, acted upon by a conservative force, is calculated to be 18.6 J. To determine the object's speed when passing through the origin, the conservation of energy principle is applied, considering both potential and kinetic energy. At x = 5.0 m, the object's kinetic energy is calculated to be 200 J, as it moves at 5.0 m/s. Since potential energy at the origin is zero, the kinetic energy remains 200 J when the object reaches x = 0. Thus, the object's speed at the origin is also 5 m/s.
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A 16 kg object is acted on by a conservative force given by F = (-2.9)x + (-4.8)x2, with F in Newtons and x in meters. Take the potential energy associated with the force to be zero when the object is at x = 0. What is the potential energy of the system associated with the force when the object is at x = 2.0 m?
I found this answer to be 18.6 J

But I can't get the second half of the question...

If the object has a velocity of 5.0 m/s in the negative direction of the x-axis when it is at x = 5.0 m, what is its speed when it passes through the origin?

I thought I could do it using the formulas for constant acceleration but I don't know the acceleration or the time. Any suggestion for what formula to use?
 
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Use the princlipple of conservation of energy. You have potential energies for both locations and a kinetic energy ao one location. The problem states the force acting on the particle is conservative so POCOE is fairly easy to implement here. Just realize that the sum of the potential and kinetic energy at 5m equals the sum of the energys at 0m.

Good luck.

[edit]Sorry, you need to use work-energy here. The force applied over a distance is work which must be factored in. Basically, PE1+KE1+W=PE2+KE2.
 
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To calculate the potential energy at x = 2.0 m, we can use the formula for potential energy:

PE = -∫F(x)dx

Since the force is given by F = (-2.9)x + (-4.8)x^2, we can integrate it to get:

PE = -∫((-2.9)x + (-4.8)x^2)dx

PE = -(-1.45x^2 - 1.6x^3) + C

Since we are given that the potential energy is zero at x = 0, we can set C = 0.

Therefore, the potential energy at x = 2.0 m is:

PE = -(-1.45(2.0)^2 - 1.6(2.0)^3) = 18.6 J

For the second part of the question, we can use the conservation of energy principle:

KE + PE = constant

At x = 5.0 m, the potential energy is zero (since we set it to be zero at x = 0) and the kinetic energy is given by:

KE = 1/2mv^2

We are given that the object has a velocity of 5.0 m/s in the negative direction of the x-axis, so we can plug in the values to get:

1/2(16)(5.0)^2 = 200 J

This is the total energy of the system at x = 5.0 m.

When the object passes through the origin (x = 0), the potential energy will still be zero, but the kinetic energy will be:

KE = 200 - PE = 200 - 0 = 200 J

To find the speed, we can plug in the values to the kinetic energy formula:

KE = 1/2mv^2

200 = 1/2(16)v^2

v = √(200/8) = 5 m/s

Therefore, the speed of the object when it passes through the origin is 5 m/s.
 
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