Work (Conservation of Energy/Newton's Laws)

[jzwiep]

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 + W_Headwind

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

Which got me= 15.7270m/s

(b)

F_net = F_g - F_headwind
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?

 

[Doc Al]

F_net = F_g - F_headwind
a=(mgsin(20)-200/cos(20))/m

What's the component of the 200N force parallel to the slope?

 

[ashishsinghal]

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

 

[jzwiep]

Thanks. I drew the triangle backwards.

 

[rwisz]

Your calculations seem to be correct. However, please note that in part (b), you have used the formula for calculating acceleration (a) instead of velocity (v). The correct formula to use would be v = v0 + at, where v0 is the initial velocity (0 m/s in this case). Using this formula, the final velocity (v) would be 15.7270 m/s, which is the same as the result obtained in part (a). This shows that both approaches, using conservation of energy and Newton's laws, yield the same result. This is because both principles are based on the fundamental concept of conservation of energy. Good job on your calculations!