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PHYS 201-002 General Physics Fall 2018 <HW Set #5: Chapter 6 Problems Problem 6.53 Part A Constants A 66-kg skier starts from rest at the top of a 1200-m-long trail which drops a total of 210 m from top to bottom. At the bottom, the skier is moving 11 m/s How much energy was dissipated by friction? Express your answer to two significant figures and include the appropriate units 1.4.10 J Submit X Incorrect, Try Again; 2 attempts remaining < Return to Assignment Provide Feedback

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Answer #1

The total energy at the top is only the gravitational potentaial energy
   Ei = mgh
And the total energy at the bottom is only the kinetic energy
   E_f=\frac{1}{2}mv^2
So, energy dissipated is
   \delta E =E_i-E_f=mgh-\frac{1}{2}mv^2
Putting the given values, we get
   \delta E =m(gh-\frac{1}{2}v^2)
\Rightarrow \delta E =66\times (9.81\times 210-\frac{1}{2}\times 11^2)
\Rightarrow \delta E =1.3\times 10^5 ~J

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