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UNO [17]
2 years ago
6

Using the law of conservation of energy, what will be the KE of an arrow having a PE of 65J after it is shot from a bow?

Physics
1 answer:
Colt1911 [192]2 years ago
7 0

Answer:

KE = 65 J (The potential energy stored in the stretched bow is converted to kinetic energy of the moving bow)

Explanation:

Law of conservation of energy states that energy can neither be created nor be destroyed but transferred from one form to another. In other words, the total energy of the system at any point is always a constant.

Now, when the bow is in stretched position, the energy associated with the stretched bow is potential energy (PE).

The energy associated with motion of the body is kinetic energy (KE).

So, the stretched bow is not moving. So, KE at the start is 0.

Therefore, total energy initially is given as:

Initial total energy = PE + KE = PE + 0 = 65 J

Now, when the bow is shot, the bow is in motion. So, the total energy of the bow is only due to kinetic energy associated with it now. Also, the potential energy of a moving bow is 0.

Therefore, final total energy is given as:

Final total energy = Final KE + Final PE = Final KE + 0 = Final KE

Now, we know, from conservation of energy,

Initial total energy = Final total energy

⇒ 65 J = KE

or Final KE = 65 J

Therefore, the KE of an arrow having a PE of 65J after it is shot from a bow is 65 J.

Hence, the potential energy stored in the stretched bow is converted to kinetic energy of the moving bow.

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Equation Faraday's law and Ohm's law we have,

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