Work (Conservation of Energy/Newton's Laws)

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Sam, a 75kg skier, descends a 50m high, 20-degree frictionless slope while facing a 200N headwind. Using work and energy principles, he calculates his speed at the bottom to be approximately 15.73 m/s, while applying Newton's laws yields a lower speed of about 12.23 m/s. The discrepancy arises from an error in calculating the force component of the headwind; it should be 200*cos(20) instead of 200/cos(20). This miscalculation affects the net force and acceleration used in the Newton's laws approach. Correcting this will align both methods and provide a consistent answer.
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Homework Statement


MP 11.44

Sam, whose mass is 75kg, straps on his skis and starts down a 50-m-high, 20degree frictionless slope. A strong headwind exerts a horizontal force of 200N on him as he skies. Find Sam's speed at the bottom (a) using work and evergy (b) Using Newton's laws


Homework Equations



v2=v02 + 2ad

Dot product

The Attempt at a Solution



(a)

U = K + WHeadwind

vf=sqrt(2*(mgh+F*(h/sin20)*cos(160))/m)

Which got me= 15.7270m/s

(b)

Fnet = Fg - Fheadwind
a=(mgsin(20)-200/cos(20))/m

v2=v02 + 2ad
v= sqrt(2ad)

Which got me: 12.225m/s

I'm not sure where I went wrong. Any thoughts?
 
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jzwiep said:
Fnet = Fg - Fheadwind
a=(mgsin(20)-200/cos(20))/m
What's the component of the 200N force parallel to the slope?
 
200/cos(20) means that component of force along the slope is greater than the force itself!
You should write 200*cos20. Well, maybe it is just a typing error (like evergy) because you wrote mg*sin20 and not mg/sin20
 
Thanks. I drew the triangle backwards.
 

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