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A 2.2 kg rock rests on the edge of a bridge that is 3.3 m above a river.What is the gravitationa l potential energy of the rock when measured from the surface of the river? What is the gravitationa potential energy when measured from the edge of the bridge?

Question

A 2.2 kg rock rests on the edge of a bridge that is 3.3 m above a river.What is the gravitationa l potential energy of the rock when measured from the surface of the river? What is the gravitationa potential energy when measured from the edge of the bridge?

A 2.2 kg rock rests on the edge of a bridge that is 3.3 m above a river.What is the
gravitationa l potential energy of the rock when measured from the surface of the
river? What is the gravitationa potential energy when measured from the edge of
the bridge?

Solution

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KaylieAdvanced · Tutor for 1 years

Answer

Here's how to calculate the gravitational potential energy (GPE) of the rock:<br /><br />**Understanding Gravitational Potential Energy**<br /><br />GPE is the energy an object possesses due to its position in a gravitational field. It's calculated using the formula:<br /><br />GPE = mgh<br /><br />Where:<br /><br />* m = mass (in kg)<br />* g = acceleration due to gravity (approximately 9.8 m/s² on Earth)<br />* h = height (in meters)<br /><br />**Calculations**<br /><br />* **GPE relative to the river:**<br /><br /> * m = 2.2 kg<br /> * g = 9.8 m/s²<br /> * h = 3.3 m<br /><br /> GPE = (2.2 kg) * (9.8 m/s²) * (3.3 m) = 71.274 J (approximately 71.3 J)<br /><br />* **GPE relative to the bridge:**<br /><br /> * m = 2.2 kg<br /> * g = 9.8 m/s²<br /> * h = 0 m (since the height is measured from the bridge itself)<br /><br /> GPE = (2.2 kg) * (9.8 m/s²) * (0 m) = 0 J<br /><br />**Explanation**<br /><br />The GPE of the rock relative to the river is 71.3 J. This means it has the *potential* to do 71.3 J of work as it falls to the river.<br /><br />The GPE of the rock relative to the bridge is 0 J. This makes sense because GPE is about *relative position*. If we're measuring from the bridge itself, the rock has no height and therefore no potential energy to fall further relative to that point.<br />
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