(11%) Problem 8: A 72.5-kg cross-country skier is climbing a 3.15° slope at a co
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(11%) Problem 8: A 72.5-kg cross-country skier is climbing a 3.15° slope at a constant speed of 1.75 m/s and encounters air resistance of 24 N 33% Part (a) Find his power output, in watts, for the work he does against the gravitational force and air resistance Grade Summary Deductions Potential 0% 100% Submissions tan() | | ( acos) sinO cosO HOME Attempts remaining: 3 % per attempt) detailed view atan0acotanO sinh0 coshO tanhO cotanhO Degrees O Radians 0 END Submit Hint I give up! Hints: 4% deduction per hint. Hints remaining: 1 Feedback: deduction per feedback. 33% Part (b) What is the magnitude of the average force, in newtons, does he exert backward on the snow to accomplish this? 33% Part (c) If he continues to exert this force and to experience the same air resistance when he reaches a level area, how long will it take him to reach a velocity of 6.5 m/s? All content © 2017 Expert TA, LLCExplanation / Answer
Try a 10 second "snapshot".
In 10 sec., the skier travels (1.75x 10) = 17.5 metres.
The height he rises is sin (3.15) x 17.5 = 0.962 metre.
The force due to gravity at the angle = (mg sin 3.15) = (72.5 x 9.8) sin 3.15 = 39.042 N.
Add 24N. wind, = 63.042 N.
a) Work = (force x distance) = (63.042 x 0.962) = 60.646 Joules.
Divide by 10s., = 6.06 Watts.
b) 63.042 N.
On the level if he still applies the same force, he will accelerate.
Just at the top of the slope, he is already doing 1.75m/sec.
Acceleration = (f/m) = (63.042/72.5), = 0.869 m/sec^2.
c) Change in V = (6.5 - 1.75) = 4.75 m/sec.
Time = (v/a) = (4.75/0.869) = 5.466 secs.
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