Potential energy of ball thrown upwards

In summary, the problem involves determining the gravitational potential energy of a 1.00 kg ball thrown upwards with an initial speed of 16.0 m/s and representing it on a graph. The horizontal line on the graph represents the total energy of the ball. For part B, the task is to draw a new graph showing the gravitational potential energy vs. height for a 2.00 kg ball with the same initial velocity. This can be represented by a line with double the slope of the original graph.
  • #1
sp3sp2sp
100
4

Homework Statement


A 1.00 kg ball is thrown directly upward with an initial speed of 16.0 m/s.
A graph of the ball's gravitational potential energy vs. height, Ug(h), for an arbitrary initial velocity is given in Part A. The zero point of gravitational potential energy is located at the height at which the ball leaves the thrower's hand.

PART A: Draw a line on the graph representing the total energy E of the ball.
For this problem, take g=10.0m/s2 as the acceleration due to gravity.

(I did this part correctly and it is the horixontal line on the graph)

PART B: Draw a new gravitational potential energy vs. height graph to represent the gravitational potential energy if the ball had a mass of 2.00 kg. The graph for a 1.00-kg ball with an arbitrary initial velocity is provided again as a reference.

Take g=10.0m/s2 as the acceleration due to gravity

2. Homework Equations

(1/2)mv^2 = -mgh


The Attempt at a Solution


the horizontal line on the attached graph is the answer to part A, which asked "Draw a line on the graph representing the total energy E of the ball."

For part B, I thought that because mass cancels in the equation that total but I then realized it asks about total gravitational PE, (=mgh) not total energy . So I am now confused how to represent that on the graph?

thanks for any help
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  • #2
There are two lines on the graph. You identified the horizontal line as the total energy in part A and that is correct. What do you think the second line represents? Hint: Read the statement of the problem.
 
  • #3
thanks for reply. according to statement it is grav PE vs height. But what confuses me is how to represent the grav PE. I mean it depends on what h and m are, so is the question asking for another PE vs height (sloped) line?
 
  • #4
sp3sp2sp said:
... so is the question asking for another PE vs height (sloped) line?
Yes, you are shown the graph of PE vs. h when the mass is 1.0 kg and you are asked to draw PE vs. h when the mass is 2.0 kg. What do you think the second plot should look like?
 
  • #5
mass doubled so same line but double the slope? (because mg = slope?)
 
  • #6
sp3sp2sp said:
mass doubled so same line but double the slope? (because mg = slope?)
Correct.
 

1. What is potential energy of a ball thrown upwards?

Potential energy is the energy an object possesses due to its position or state. In the case of a ball thrown upwards, it is the energy it has due to its height above the ground.

2. How is the potential energy of a ball thrown upwards calculated?

The potential energy of a ball thrown upwards can be calculated using the formula: PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the ball above the ground.

3. Does the potential energy of a ball thrown upwards change?

Yes, the potential energy of a ball thrown upwards changes as it moves. As the ball reaches its maximum height, the potential energy is at its highest. As it falls back to the ground, the potential energy decreases.

4. What factors can affect the potential energy of a ball thrown upwards?

The potential energy of a ball thrown upwards can be affected by the mass of the ball, the height it is thrown from, and the acceleration due to gravity. Other factors such as air resistance and wind can also have an impact on the potential energy.

5. How is potential energy of a ball thrown upwards related to kinetic energy?

Potential energy and kinetic energy are two forms of energy that are related to each other. As the ball is thrown upwards, it gains potential energy. As it falls back to the ground, the potential energy is converted into kinetic energy, which is the energy of motion.

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