Evaluate the solution A student was given the following...

Evaluate the solution A student was given the following problem: A 0.20 -kg cannonball with a $+1.0 \times 10^{-4} \mathrm{C}$ charge starts at rest 0.40 $\mathrm{m}$ from a fixed $+2.0 \times 10^{-4} \mathrm{C}$ charge. When the cannonball is released, it flies vertically upward. How fast is it moving when 10 m above the fixed charge? The student gave this solution: $$\left(9.0 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right)\left(+2.0 \times 10^{-4} \mathrm{C}\right)\left(+1.0 \times 10^{-4} \mathrm{C}\right) /(0.40 \mathrm{m})^{2}$ $=(1 / 2)(10 \mathrm{kg}) \mathrm{v}_{y}^{2}+(0.20 \mathrm{kg})\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(10 \mathrm{m})$$ or $v_{y}=15 \mathrm{m} / \mathrm{s}$ . Evaluate the solution to this problem and correct any errors you find.

Evaluate the solution A student was given the following problem: A 0.20 -kg cannonball with a $+1.0 \times 10^{-4} \mathrm{C}$ charge starts at rest 0.40 $\mathrm{m}$ from a fixed $+2.0 \times 10^{-4} \mathrm{C}$ charge. When the cannonball is released, it flies vertically upward. How fast is it moving when 10 m above the fixed charge? The student gave this solution:
$$\left(9.0 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right)\left(+2.0 \times 10^{-4} \mathrm{C}\right)\left(+1.0 \times 10^{-4} \mathrm{C}\right) /(0.40 \mathrm{m})^{2}$
$=(1 / 2)(10 \mathrm{kg}) \mathrm{v}_{y}^{2}+(0.20 \mathrm{kg})\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(10 \mathrm{m})$$
or $v_{y}=15 \mathrm{m} / \mathrm{s}$ . Evaluate the solution to this problem and correct any errors you find.

Key Concepts

Conservation of Energy

The principle of conservation of energy states that the total energy in an isolated system remains constant. In physics problems, especially those involving forces that do work (such as conservative forces like gravity or the electrostatic force), one must account for all forms of energy—kinetic and potential—making sure that the change in potential energy equals the gain in kinetic energy (or vice versa) as the object moves. This ensures that energy is neither created nor destroyed but only transformed from one form to another.

Electrostatic Potential Energy

Electrostatic potential energy is the energy stored between charged objects due to their positions relative to each other. It is usually determined by Coulomb's law, which shows that the potential energy is proportional to the product of the two charges and inversely proportional to the distance between them. When the distance or configuration of charges changes, a corresponding change in this energy occurs, which can be converted into kinetic energy or other forms of energy.

Gravitational Potential Energy

Gravitational potential energy is associated with an object's position within a gravitational field. The most common expression for this energy is mgh, where m is mass, g is the acceleration due to gravity, and h is the height above a defined reference point. In energy conservation problems that involve vertical motion, it is crucial to accurately calculate the change in gravitational potential energy as the object moves, as it directly impacts the amount of kinetic energy the object can gain or lose.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion and is calculated with the formula 1/2 mv², where m is the mass and v is the velocity. In problems utilizing energy conservation, the increase in kinetic energy of an object is balanced by the loss in potential energy, assuming no non-conservative forces (like friction) are doing work. Accurate use of this concept is necessary for correctly determining changes in motion based on energy transformations.

Transcript

00:01 Okay, so the question was asking us to determine how fast is it moving when 10 meter above the fixed charge.

00:07 Well, we know in order to find how fast the charge was moving, we need to determine the work.

00:17 In order to determine the work, we've got to find the changing potential energy.

00:22 And to find a changing potential energy, we have to know the initial potential energy and the final potential energy.

00:27 So for the initial potential energy is equal to kq2 over rri.

00:31 K is k1 k2 is two different charges.

00:34 R is the initial distance, which is given 0 .4 meter, okay? and the final potential energy is the final electric potential energy plus the potential energy that was gain when the charge was moved up, okay? so, which is equal to kq1 over rf plus mgh.

00:56 Rf is the final distance.

00:58 H is the height from from the starting point to the final point, which is from 0 .4 meter to 10 meter.

01:09 Okay? so we know that the work done is equal to negative changing potential energy, which is equal to negative uf minus ui, which is equal to ui minus uf.

01:20 And we know the work can be equal to the kinetic energy in this case, which is 1 1 .5 mv squared.

01:26 Therefore 1 .5 mv squared is equal to kq1 .2 over i minus k k kk1 kk2 over rf minus m g.

01:34 So the equation that was given from the question is not complete.

01:38 Why is that? because it's correct on the left side.

01:42 On the left side was just simply the initial potential energy, or in other words, the initial electric potential energy, because at the beginning there's only electric potential energy.

01:52 So the electric potential energy initial, in this case, is equal to the initial potential energy.

01:58 But on the right side, it was only given the work or the kinetic energy for this question...

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