Potential energy and work relationship

[cookiemnstr510510]

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=U_f-U_i, 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

 

[SammyS]

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=U_f-U_i, 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.

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.

 

[cookiemnstr510510]

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.

 

[cookiemnstr510510]

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=mgh_f-mgh_i the "h's" are not vector quantities therefore they just add. but for W=Fdcosθ, the "d" is a vector quantity.

 

[SammyS]

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 $\ \hat j \,.$

 

[SammyS]

Now, let's look at the following part of the Original Post of this thread.

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=U_f-U_i, 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 h_i = 10m to h_f = 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.

 

[cookiemnstr510510]

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.