Potential energy and work relationship

In summary: So if ΔU=-mghf-mghi and Work=-Fdcosθ then ΔU=-mghi+Fdcosθ. Okay so now that we have that, does this mean that if I have two masses and I want to find the work done on them by gravity, I have to do the following:ΔU=-mghi+Fdcosθ
  • #1
cookiemnstr510510
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Homework Statement


Taking my Electricity and Magnitism course right now and we are in the section in the book of the electric potential.
I have some basic questions regarding work and potential energy.
Lets say we are holding a ball 10m above the ground and release it. I am defining the ground to be zero.
The ball has all potential and no kinetic energy when it is in my hand at 10m.
If i want to find the potential energy of this system once I drop the ball and it touches the ground is:
ΔU=Uf-Ui, this will give me a negative ΔU, since the final "h" of the ball is 0m and initial "h" of the ball is 10m. I am fine with this.
Now when thinking about the work, work is done on the ball by the force (forge of gravity)
work is defined: W=Fdcosθ, the "d" is displacement.
Why is the displacement not negative? is the displacement ever negative? is a better way to think about this making "d" distance rather than displacement since a distance cannot be negative?

Homework Equations


above

The Attempt at a Solution


above
 
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  • #2
cookiemnstr510510 said:

Homework Statement


Taking my Electricity and Magnitism course right now and we are in the section in the book of the electric potential.
I have some basic questions regarding work and potential energy.
Lets say we are holding a ball 10m above the ground and release it. I am defining the ground to be zero.
The ball has all potential and no kinetic energy when it is in my hand at 10m.
If i want to find the potential energy of this system once I drop the ball and it touches the ground is:
ΔU=Uf-Ui, this will give me a negative ΔU, since the final "h" of the ball is 0m and initial "h" of the ball is 10m. I am fine with this.
Now when thinking about the work, work is done on the ball by the force (forge of gravity)
work is defined: W=Fdcosθ, the "d" is displacement.
Why is the displacement not negative? is the displacement ever negative? is a better way to think about this making "d" distance rather than displacement since a distance cannot be negative?

Homework Equations


above

The Attempt at a Solution


above
Let's address the following part of your post. It appears to be the main question you actually are asking at this point, although it looks like it leads to other question you have in mind.
cookiemnstr510510 said:
Now when thinking about the work, work is done on the ball by the force (forge of gravity)
work is defined: W=Fdcosθ, the "d" is displacement.
Why is the displacement not negative? is the displacement ever negative? is a better way to think about this making "d" distance rather than displacement since a distance cannot be negative?
If you are considering the work done on the ball by gravity, then both the force and the displacement have the same direction. If we consider 1-Dimensional motion, in which case we can consider the directions of vectors to be given by using signs, then both the force and the displacement are negative. Their product is positive.

In general displacement is a vector quantity.
 
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  • #3
SammyS said:
Let's address the following part of your post. It appears to be the main question you actually are asking at this point, although it looks like it leads to other question you have in mind.

If you are considering the work done on the ball by gravity, then both the force and the displacement have the same direction. If we consider 1-Dimensional motion, in which case we can consider the directions of vectors to be given by using signs, then both the force and the displacement are negative. Their product is positive.

In general displacement is a vector quantity.
Okay, so really force is a vector quantity in the negative j-hat direction and so is displacement so
W=F(-jhat)d(-jhat)cosθ=Fdcosθ because <0,-1,0>⋅<0,-1,0>=<0,1,0> which is positive jhat.
 
  • #4
SammyS said:
Let's address the following part of your post. It appears to be the main question you actually are asking at this point, although it looks like it leads to other question you have in mind.

If you are considering the work done on the ball by gravity, then both the force and the displacement have the same direction. If we consider 1-Dimensional motion, in which case we can consider the directions of vectors to be given by using signs, then both the force and the displacement are negative. Their product is positive.

In general displacement is a vector quantity.
So the difference between ΔU and Work is for ΔU=mghf-mghi the "h's" are not vector quantities therefore they just add. but for W=Fdcosθ, the "d" is a vector quantity.
 
  • #5
cookiemnstr510510 said:
Okay, so really force is a vector quantity in the negative j-hat direction and so is displacement so
W=F(-jhat)d(-jhat)cosθ=Fdcosθ because <0,-1,0>⋅<0,-1,0>=<0,1,0> which is positive jhat.
Well, not exactly.

Work is the scalar product of two vector quantities, force and displacement.

That can be expressed as ##\ W = \vec{F} \cdot \vec{d} \ ## and equivalently as ## W = | \vec{F}| \, |\vec{d}|\, \cos(\theta) \,, \ ## where ##\ \theta \ ## is the angle from the direction of ##\ \vec F \ ## to the direction of ##\ \vec d \ ##.

Also:
##<0, -1, 0> \cdot <0, -1, 0>\ =\ (0)(0)+(-1)(-1)+(0)(0) = -1\ ##, a scalar, not ##\ \hat j \,.##
 
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  • #6
Now, let's look at the following part of the Original Post of this thread.
cookiemnstr510510 said:
1
Lets say we are holding a ball 10m above the ground and release it. I am defining the ground to be zero.
The ball has all potential and no kinetic energy when it is in my hand at 10m.
If i want to find the potential energy of this system once I drop the ball and it touches the ground is:
ΔU=Uf-Ui, this will give me a negative ΔU, since the final "h" of the ball is 0m and initial "h" of the ball is 10m. I am fine with this.
Here ΔU is the change in potential energy, which is negative, as you point out. This is due to Δh being negative.
Also notice, immediately before the ball touches the ground, ΔKE, the change in kinetic energy is equal to −ΔU, which is positive since ΔU is itself negative.

In fact, ΔKE = −ΔU = −(mg)(Δh), remembering that Δh is negative
For gravity acting on a ball of mass, m, the force of gravity is −(mg), the negative sign used for the downward direction, and the displacement is Δh, which is also negative. This gives us that ΔKE here is indeed equal to the work, W, done on the ball by gravity. in falling from hi = 10m to hf = 0m . The fact that the change in kinetic energy, ΔKE, is equal to the work done on the ball, W, is consistent with the Work-Energy Theorem.

If you look carefully at how a potential energy function is defined, it's not in terms of the work done by the gravitational force, or the the electric force as a mass or charge is displaced. Potential energy is defined in terms of the work done by an external force ( which opposes the gravitational or electric force ) in displacing the mass or charge.

So, it's no surprise that there is this issue of opposite signs.
 
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  • #7
SammyS said:
If you look carefully at how a potential energy function is defined, it's not in terms of the work done by the gravitational force, or the the electric force as a mass or charge is displaced. Potential energy is defined in terms of the work done by an external force ( which opposes the gravitational or electric force ) in displacing the mass or charge.
Ahh, okay this is making a bit more sense.
 

FAQ: Potential energy and work relationship

1. What is potential energy and how is it related to work?

Potential energy is the energy that an object possesses due to its position or configuration. It is related to work through the work-energy theorem, which states that the change in an object's potential energy is equal to the work done on the object. In other words, when work is done on an object, its potential energy changes by the same amount.

2. How is potential energy calculated?

The formula for calculating potential energy depends on the type of potential energy being considered. For gravitational potential energy, the formula is PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object. For elastic potential energy, the formula is PE = 1/2kx², where k is the spring constant and x is the displacement from equilibrium.

3. Can potential energy be negative?

Yes, potential energy can be negative. This typically occurs when the reference point for potential energy is set at a higher point than the object's actual position. For example, if the reference point is set at the top of a hill and the object is at the bottom of the hill, its potential energy would be negative.

4. How does potential energy affect an object's motion?

Potential energy does not directly affect an object's motion. However, potential energy can be converted into kinetic energy, which does affect an object's motion. For example, when an object is dropped from a height, its potential energy is converted into kinetic energy as it falls, increasing its speed.

5. Can potential energy be converted into other forms of energy?

Yes, potential energy can be converted into other forms of energy. For example, gravitational potential energy can be converted into kinetic energy as an object falls. Elastic potential energy can also be converted into kinetic energy when a spring is released. Additionally, potential energy can be converted into thermal energy through friction or into electrical energy through certain types of generators.

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