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The gravitational potential energy of a 10 kg boulder is 1024J. What is its PE at 50.0 m abovethe ground? At 20.0 m? at 1.00 m? On the ground? Make a graph of height versus energy.What is the shape of your graph?

Question

The gravitational potential energy of a 10 kg boulder is 1024J. What is its PE at 50.0 m abovethe ground? At 20.0 m? at 1.00 m? On the ground? Make a graph of height versus energy.What is the shape of your graph?

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Solution

To solve this problem, we first need to understand the formula for gravitational potential energy (PE), which is PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height above the ground.

Given that the mass (m) of the boulder is 10 kg and the gravitational potential energy (PE) is 1024 J, we can rearrange the formula to solve for g (acceleration due to gravity): g = PE / (m * h).

  1. First, we need to find the value of g. We know that the PE is 1024 J when the boulder is at an unknown height. So, we have 1024 J = 10 kg * g * h. Unfortunately, we don't have the value for h, so we can't find the exact value for g in this case.

However, on Earth, the acceleration due to gravity is approximately 9.8 m/s². We can use this value for g in the following calculations.

  1. To find the PE at 50.0 m above the ground, we substitute these values into the formula: PE = 10 kg * 9.8 m/s² * 50.0 m = 4900 J.

  2. To find the PE at 20.0 m above the ground, we do the same: PE = 10 kg * 9.8 m/s² * 20.0 m = 1960 J.

  3. To find the PE at 1.00 m above the ground: PE = 10 kg * 9.8 m/s² * 1.00 m = 98 J.

  4. On the ground, the height (h) is 0, so the PE is also 0, because PE = mgh = 10 kg * 9.8 m/s² * 0 = 0 J.

To make a graph of height versus energy, you would plot the height on the x-axis and the energy on the y-axis. You would then plot the points (50.0 m, 4900 J), (20.0 m, 1960 J), (1.00 m, 98 J), and (0, 0). The shape of the graph would be a straight line starting from the origin (0,0) and going upwards, showing that the potential energy increases linearly with height.

This problem has been solved

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