# How to find the average force opposing motion down a slope? [closed]

A skier of mass 75 kg accelerates from rest down the slope as shown in the diagram. At the bottom of the slope, he is traveling at 40 m/s. What is the average force opposing his motion down the slope? (Use g = 10 N/kg)

This is what I've tried:

P.E at the top = 75 * 10 * 300 = 225 000 J

K.E at the bottom = (1/2) * m * v^2 = 0.5 * 75 * 40^2 = 60000 J

Energy lost in friction: 225000 - 60000 =165000 J (work done in opposing motion)

Work done = force X distance 165000 = f X 500

165000/500 = f

f = 330 N

Is that correct?

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What have you tried? Is this a homework question? – tpg2114 Dec 17 '12 at 11:12
I tried solving it.. but I don't know what approach to use. It's not a homework question.. I was just revising. Could you help me solve it? – John Dec 17 '12 at 11:14
This looks like a homework problem, and the site rules forbid answering such questions. However, you can work out the answer from conservation of energy. The work done by friction is the frictional force times distance moved, and if you add this to the skier's kinetic energy the sum has to equal the change in the skier's potential energy. – John Rennie Dec 17 '12 at 11:14
I tried solving it.. is it correct? Mass (m) = 75 kg. Acceleration due to gravity (g) = 9.8 (m/s)/s. The slope angle = inv tan ( 300 / 400 ) = 36.87 °. Initial velocity (u) = 0 m/s. Final velocity (v) = 40 m/s. distance (s) travelled down the slope = 500 meters. The force down the slope on the skier due to gravity = m * g * sin 36.87 ° = 441 Newtons. The acceleration (a) down the slope = v² / ( 2 * s ) = 1600 / 1000 = 1.6 (m/s)/s The force required to accelerate 75 kg @ 1.6 (m/s)/s = 75 * 1.6 = 120 Newtons The force opposing motion = 441 - 120 = 321 Newtons – John Dec 17 '12 at 11:16
Hi, John, and welcome to Physics Stack Exchange! Please see our homework policy. We expect homework problems to have some effort put into them, and deal with conceptual issues. If you edit your question to explain (1) What you have tried, (2) the concept you have trouble with, and (3) your level of understanding, I'll be happy to reopen this. (Flag this message for ♦ attention with a custom message, or reply to me in the comments with @Manishearth to notify me) – Manishearth Dec 17 '12 at 12:38