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Ksenya-84 [330]
3 years ago
13

A block with a mass of 8.7 kg is dropped from rest from a height of 8.7 m, and remains at rest after hitting the ground. 1)If we

consider the system to consist of the block, the ground, and the surrounding air, what is the change in the internal energy of the system
Physics
1 answer:
Harlamova29_29 [7]3 years ago
7 0

To solve this problem we will apply the concepts related to gravitational potential energy.

This can be defined as the product between mass, gravity and body height.

Mathematically it can be expressed as

\Delta P = mgh

\Delta P = (8.7)(9.8)(3)

\Delta P = 255.78J

Therefore the change in the internal energy of the system is 255.78

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A robin in flight has 20.8 J of PE when it is 27.6 m high. What is the mass of the robin? (Unit = kg)
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Answer:

<h2>\boxed{  \bold{ \purple{0.0769 \: kg \: }}}</h2>

Explanation:

\sf{Potential   Energy ( P.E ) \:  =  \: 20.8 \: joule}

\sf{distance \:  = 27.6 \: metre}

\sf{mass = } ?

\sf{acceleration \: due \: to \: gravity = 9.68 \:  {metre \: per \: second}^{2} }

Now, let's find the mass:

\sf{PE  \:  =  mass \times gravity \:  \times  \: distance}

plug the values

⇒\sf{20.8 = m \times 9.8 \times 27.6}

Multiply the numbers

⇒\sf{20.8 = 270.48 \: m}

Swap the sides of the equation

⇒\sf{270.48m = 20.8}

Divide both sides of the equation by 270.48

⇒\sf{ \frac{270.48m}{270.48}  =  \frac{20.8}{270.48} }

Calculate

⇒\sf{0.0769} kg

Hope I helped!

Best regards!!

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