Understanding the Work-Energy Theorem

[GrandMaster87]

Hi

I just can't seem to grasp this concept!

Anyone have like a small tutorial for this section or notes on it?

Im in grade 12..

For example a question from the exam...

A person skis down a 20m long snow slope which makes an angle of 25degrees with the horizontal.

The total mass of the skier and skis is 50kg. There is a constant frictional force of 60N opposing the skiers motion. The speed of the skier as he descends from the top of the slope is 2.5m/s

1) Calculate the magnitude of the net force parallel to the slope experience by the person
2) Calculate the maximum spped of the skier at the bottom of the 20m slope.

Does anyone know where i can get notes on this section? i really struggle with this...i just guess i need more examples and steps on how to do this...

Thanks guys

 

[Doc Al]

What does your textbook have to say on this? If you show what you've been able to figure out so far, then we can provide useful help.

You may want to read this tutorial (especially lesson 2): http://www.physicsclassroom.com/Class/energy/index.cfm"

 

[GrandMaster87]

What does your textbook have to say on this? If you show what you've been able to figure out so far, then we can provide useful help.

You may want to read this tutorial (especially lesson 2): http://www.physicsclassroom.com/Class/energy/index.cfm"

Ok...basically this is what i have done so far.

Fgravity=mg(sinx)
=50 x 9.8 x sin25
= 207.08N
Ffriction=60N

Therefore force parallel
=207.08-60
=147.08

But they also tell us to work out work done we must use F(cosx)$$\Delta$$x

so i want to know...when and why are we using sinx here? they don't give us any explanation as to why to use the sin and i don't even know when and how to use it...

 

[Doc Al]

Ok...basically this is what i have done so far.

Fgravity=mg(sinx)
=50 x 9.8 x sin25
= 207.08N
Ffriction=60N

Therefore force parallel
=207.08-60
=147.08

Perfect!

But they also tell us to work out work done we must use F(cosx)$$\Delta$$x

Right.

so i want to know...when and why are we using sinx here? they don't give us any explanation as to why to use the sin and i don't even know when and how to use it...

You used sin25 because you needed to get the component of gravity (which acts down) parallel to the incline. (If you are fuzzy on finding components, read this: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm)

In the formula for work, the angle is the angle between the force and the displacement (nothing directly to do with the angle of the incline). Since you found the force parallel to the incline, and the displacement is along the incline, what must cosx be?

 

[GrandMaster87]

In the formula for work, the angle is the angle between the force and the displacement (nothing directly to do with the angle of the incline). Since you found the force parallel to the incline, and the displacement is along the incline, what must cosx be?

Wont cosx be Cos0 ?

 

[Doc Al]

Wont cosx be Cos0 ?

Exactly. And cos0 equals what?

 

[GrandMaster87]

Exactly. And cos0 equals what?

1.

thanks a lot :) forum is quite cool..im gng to study now...and will come back if i need more help thank you a lot Doc Al