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An object with a mass of 100 kg is dropped from a height of 20 m. If the velocity of the object before hitting the ground is 15 m/s, is there a loss of energy in the form of heat? If so, how much? Assume, g= 9.8 m/s^2

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daigz92

Answer:

Yes, there is a loss of 8350 J of energy in the form of heat

Explanation:

The principle of energy conservation is described mathematically

as the energy conservation equation as follows:

                            ΔK+ΔU=Q+W        

where:

  • ΔK: changes in kinetics energy [J]
  • ΔU: changes in potential energy [J]
  • Q: gain or loss of heat[J]
  • W: work done by the system or in the system[J]

Analyzing each term of the equation:

   [tex]ΔK=\frac{1}{2}mv_{f} ^{2}  - \x]frac{1}{2}mv_{i} ^{2}=\frac{1}{2}*100*15^{2} - \frac{1}{2}*100*0^{2}=11250 J[/tex]

ΔU=[tex]mgh_{f}-mgh_{i}=100*9,8*0-100*9,8*20=-19600[J][/tex]

Q=?

W=0 [J]

Replacing in the main equation:

11250-19600=Q+0

Q= -8350 [J]

So, the answer is YES, there is a loss of 8350 J of  energy in the form of heat.

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