Work/Energy - Incline Plane w/ Friction

[De_Dre01]

Homework Statement

A 74.5 kg snowboarder heads down a 16.0° hill that has a height of 78.4 m. If the hill is assumed to be frictionless and there is horizontal wind with a force of 93 N acting against the snowboarder, find the speed of the snowboarder as they reach the bottom of the hill using work and energy.

Homework Equations

W = FΔd
Ep = mgΔy
Ek = 1/2mv²

The Attempt at a Solution

So far, I tried calculating potential energy at top:
[/B]
Ek = mgΔy
= (74.5)(9.8)(78.4)
= 57,239.84 J

Work done by wind:
W = FΔd
= (93N)(78.4/sin16.0°)
= 26,452.14753 J

Then, I subtracted work done by wind from potential energy, then used the kinetic energy to solve for final velocity.
What am I doing wrong?

 

[TSny]

Hello, and welcome to PF!
Can you explain why you used the sine of the angle when finding Δd?

 

[De_Dre01]

Opposite = 78.4 m
Angle = 16.0°Solving for hypothenuse:
sin16° = 78.4/h
h = 78.4/sin16°

Is this not correct?

 

[TSny]

It's correct if Δd represents the distance along the slope. But, then, is the formula W = FΔd correct? What is the direction of the wind force?

 

[De_Dre01]

Horizontal wind force opposing the skateboarder's motion.

 

[TSny]

So, the force of the wind is not parallel to the displacement Δd. How do you find the work when the force is not parallel to the displacment?

 

[De_Dre01]

I must consider the incline plane, should I not?

Would I need to find the adjacent side?

Edit: Would the distance be 78.4/tan16?

 

[TSny]

In general, how do you find the work done by a force that is not parallel to the displacement?

 

[De_Dre01]

W = FΔdcosθ.

 

[TSny]

Right. Can you apply that to this problem? What would you use for θ? Use a carefully drawn diagram.

 

[De_Dre01]

74°?

 

[TSny]

No. What does θ stand for in the formula W = FΔdcosθ?

 

[De_Dre01]

The angle between the Force and Distance. Would it be 16° because of the Z Pattern?

 

[TSny]

The angle between the Force and Distance.

Yes

Would it be 16° because of the Z Pattern? Or 90°

I'm not sure what you mean by the Z pattern. Did you draw vectors representing the wind force and the displacement? Does the angle between the force and the displacement look like it's less than 90^o or greater than 90^o?

 

[De_Dre01]

I might be picturing the problem incorrectly. The wind would be directed west while displacement would be down the hill, correct?

 

[TSny]

Yes, if west means to the left in your figure back in post #3.

 

[De_Dre01]

The angle would be less than 90 then.

 

[TSny]

When getting the angle between two vectors, you should draw the two vectors from the same point. That is, the "tails" of the vectors should be at the same point.

 

[De_Dre01]

Even if you do that, can't you move the vector around and displace the tail of the second vector to the head of the first vector? Resulting in the same angle of 16°*?

 

[TSny]

No. It is easiest to see the angle between two vectors when they are drawn from the same point, as shown in the top figure here
http://hyperphysics.phy-astr.gsu.edu/hbase/work2.html

 

[De_Dre01]

I thought you could move the vector around. In that case, would the angle be 90+74°?

 

[TSny]

Yes, good.

 

[De_Dre01]

So now I have:

W = FΔdcosθ
= (93N)(78.4)cos(164)
= -7008.75 J

Correct?

 

[TSny]

Why did you use 78.4 m for Δd?

 

[De_Dre01]

Ok I calculated distance in the x by tan16 = 78.4/x now apply work formula?

 

[TSny]

OK. That should work (no pun intended). But what formula are you now going to use for the work?

 

[De_Dre01]

Ahah. Got it now. Thanks, was really helpful.