Calculating Height Using the Work-Energy Theorem

[ehabmozart]

Homework Statement

At the base of a frictionless icy hill that rises at 25 degree above horizontal, a toboggan has a speed of 12m/s toward the hill. How high vertically above the base will it go before stopping?

Homework Equations

WORK ENERGY THEOREM

The Attempt at a Solution

I know the answer but the steps were quite confusing.
W= Delta K. Where W should be the NET work. Weight does a work at 180 from the displacement. So Wmg = -mgh.. Fine! Now, the book equated this with the kinetic energy. Isn't there a normal force acting too. Why didn't the book consider the work done by the normal force. I feel i should add W done by N= (mgcos theta) x cos theta x h ... Why can't i do that... I need good replies. Thanks to whoever contributes with good answers. :)

 

[TSny]

Careful. The displacement vector of the taboggan is in the direction of the slope of the hill; that is, at 25 degrees above the horizontal. But the force of gravity is vertically downward. So, the angle between the weight and the displacement is not 180 degrees.

 

[ehabmozart]

But now, he is asking me about the vertical height. So shouldn't we think of the y-axis as a reference now?

 

[TSny]

I think it's best to stick with the definition of work as W = Fdcosθ

If you can find the distance d along the hill, then you can use that to find the height h.

So, what would be the value of θ for the weight force and what would be the value of θ for the normal force?

 

[ehabmozart]

Ok... I get your point and let me solve it with your convention. W done by gravity along s (which is the hypotanious) is equal to = -mgcos theta X s = - .5 mv^2 ... s would be h/sin theta... The answers turns out to be wrong because over all we equate and have - mg h/tan theta = .5 m v^2 ... Kindly Clarify!

 

[TSny]

Please specify what you used for the value of θ in the expression mgscosθ for the work done by the weight force.

 

[ehabmozart]

Oh my godness... I knew where my mistake is... It is W=F.s ... So we take mg as a whole , s as a whole and the angle between them is sin theta... We then mltiply it by h/sin theta .. so sin theta cancels... WOW! I owe you bro. Cs i didn't wan't to solve it with potential energy rather than with kinetic energy.. Thanks!