Question

A 66-kg skier starts from rest at the top of a 1200-m-long trail which drops a total of 170 m from top to bottom. at the bottom, the skier is moving 11 m/s.

Answer 1

A total of 1.1 × 10⁵Joules energy was dissipated by friction

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Further explanation

Let's recall Kinetic Energy Formula as follows:

$Ek = \frac{1}{2}mv^2$

Ek = Kinetic Energy ( Joule )

m = mass of the object ( kg )

v = speed of the object ( m/s )

Let us now tackle the problem !

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Complete Questions:

A 66-kg skier starts from rest at the top of a 1200-m long trail which drops a total of 170 m from top to bottom. at the bottom, the skier is moving 11.0 m/s. How much energy was dissipated by friction?

Given:

mass of skier = m = 66 kg

height of trail = h = 170 m

velocity of skier = v = 11.0 m/s

Asked:

work done by friction = W = ?

Solution:

$W = Ep - Ek$

$W = mgh - \frac{1}{2}mv^2$

$W = 66(9.8)(170) - \frac{1}{2}(66)(11.0)^2$

$W = 109956 - 3993$

$W = 105963$ $\texttt{ J}$

$W \approx 1.1 \times 10^5 \texttt{ J}$

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Conclusion:

A total of 1.1 × 10⁵ Joules energy was dissipated by friction

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441
• Newton's Law of Motion: brainly.com/question/10431582
• Example of Newton's Law: brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

Answer 2

I believe this question ask for the energy dissipated by friction.

The overall energy equation for this is:

F = PE – KE

where F is friction loss, PE is potential energy = m g h, KE is kinetic energy = 0.5 m v^2

 

F = 66 kg * 9.8 m/s^2 * 170 m – 0.5 * 66  kg * (11 m/s)^2

F = 105,963 J  ~  106,000 J