For instance, a 50 kg mass held at a height of 4 m from the ground has a gravitational potential energy of:
The most important thing to remember is that the higher an object is off the ground, the greater its gravitational potential energy.
Mechanical Energy
We now have equations relating work to both kinetic and potential energy:
Combining these two equations gives us this important result:
Or, alternatively,
As the kinetic energy of a system increases, its potential energy decreases by the same amount, and vice versa. As a result, the sum of the kinetic energy and the potential energy in a system is constant. We define this constant as E, the mechanical energy of the system:
This law, the conservation of mechanical energy, is one form of the more general law of conservation of energy, and it’s a handy tool for solving problems regarding projectiles, pulleys, springs, and inclined planes. However, mechanical energy is not conserved in problems involving frictional forces. When friction is involved, a good deal of the energy in the system is dissipated as heat and sound. The conservation of mechanical energy only applies to closed systems.
Example 1
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A student drops an object of mass 10 kg from a height of 5 m. What is the velocity of the object when it hits the ground? Assume, for the purpose of this question, that g = –10 m/s2. | |
Before the object is released, it has a certain amount of gravitational potential energy, but no kinetic energy. When it hits the ground, it has no gravitational potential energy, since h = 0, but it has a certain amount of kinetic energy. The mechanical energy, E, of the object remains constant, however. That means that the potential energy of the object before it is released is equal to the kinetic energy of the object when it hits the ground.
When the object is dropped, it has a gravitational potential energy of:
By the time it hits the ground, all this potential energy will have been converted to kinetic energy. Now we just need to solve for v:
Example 2
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Consider the above diagram of the trajectory of a thrown tomato: |
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At what point is the potential energy greatest? |
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At what point is the kinetic energy the least? |
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At what point is the kinetic energy greatest? |
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At what point is the kinetic energy decreasing and the potential energy increasing? |
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At what point are the kinetic energy and the potential energy equal to the values at position A? | |
The answer to question 1 is point B. At the top of the tomato’s trajectory, the tomato is the greatest distance above the ground and hence has the greatest potential energy.
