A 20 kg object is acted on by a conservative force given by F = -3.0x - 5.0x^2,

Question

A 20 kg object is acted on by a conservative force given by F = -3.0x - 5.0x^2, 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. (a) What is the potential energy of the system associated with the force when the object is at x = 6.0 m? 414 J (b) If the object has a velocity of 6.0 m/s in the negative direction of the x axis when it is at x = 8.0 m, what is its speed when it passes through the origin? 11.443 m/s (c) What are the answers to (a) and (b) if the potential energy of the system is taken to be -7.0 J when the object is at x = 0? (a) J (b) m/s.

Explanation / Answer

F = -3x -5x2

We know F = -dU/dx

So -dU/dx = -3x -5x2

or dU = (3x + 5x2)dx

Let us integrate above with limit for x from 0 to x

So we have U(x) - U(0) = 3x2/2 + 5x3/3 --------------- (1)

Where U(x) is the potential at x and U(0) is the potential at x = 0.

But U(0) = 0, given

So we have U(x) =  3x2/2 + 5x3/3

a) we have to find U(6.0m)

So U(6.0m) = 3(6m)2/2 + 5(6m)3/3 = 414J

b) potential energy at x = 8.0m is

U(8.0m) = 3(8m)2/2 + 5(8m)3/3 = 949.333J

And Kinetic energy at x = 8.0m, where speed is 6m/s, is:

KE = (1/2)mv2 = (1/2)(20kg)(6m/s)2 = 360J

So, total mechanical energy at x = 8.0m is:

E = U(8.0m) + KE = 949.333J + 360J = 1309.333J

Now, since the force is conservative, the total mechanical energy would remain constant throughout the motion of the object.

So, total mechanical enery at x = 0 will also be E = 1309.333J

At x = 0, potential energy is 0.

At x = 0 Kinetic energy is = 1/2mv02, where v0 is the speed at x = 0.

So, at x = 0, mechanical energy = potential energy + Kinetic energy = 0 + (1/2)mv02 = (1/2)mv02

mechanical energy at x = 0 is equal to mechanical energy at x = 8m

or (1/2)mv02 = E = 1309.333J

or (1/2)(20kg)v02 = 1309.333J

or v0 = 11.4426m/s

(c) let us return to equation (1) above:

U(x) - U(0) = 3x2/2 + 5x3/3

Now we have U(0) = -7J

So U(x) =  3x2/2 + 5x3/3 - 7

So potential energy at x = 6m will be

U(6) = 3(6m)2/2 + 5(6m)3/3 - 7J

or U(6) = 414J - 7J = 407J

Now we have to find speed at x = 0

So, potential energy at x = 8.0m is

U(8.0m) = 3(8m)2/2 + 5(8m)3/3 - 7J = 942.333J

And Kinetic energy at x = 8.0m, where speed is 6m/s, is:

KE = (1/2)mv2 = (1/2)(20kg)(6m/s)2 = 360J

So, total mechanical energy at x = 8.0m is:

E(8m) = U(8.0m) + KE = 942.333J + 360J = 1302.333J

at x = 0, mechanical energy = potential energy + Kinetic energy = 0 + (1/2)mv02 = (1/2)mv02

So total mechanical energy at x = 0 is:

E(0) = (1/2)mv02

since the force is conservative, the total mechanical energy would remain constant throughout the motion of the object. So,

E(0) = E(8m)

or (1/2)mv02 = 1302.333J

or (1/2)(20kg)v02 = 1302.333J

or v0 = 11.4119m/s

So, speed at x = 0 will be v0 = 11.4119m/s.

This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification I will be happy to oblige....

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