Understanding the Work-Energy Theorem

In summary, the conversation is about a student in grade 12 struggling with a physics concept and asking for notes or a tutorial. Another user suggests utilizing a physics tutorial website and provides a link. The student then shares their progress and asks for clarification on when to use sin and cos in the equations. The expert explains the reasoning behind using sin and cos in the given scenario and encourages the student to continue studying.
  • #1
GrandMaster87
18
0
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
 
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  • #2
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"
 
Last edited by a moderator:
  • #3
Doc Al said:
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)[tex]\Delta[/tex]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...
 
Last edited by a moderator:
  • #4
GrandMaster87 said:
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)[tex]\Delta[/tex]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?
 
  • #5
Doc Al said:
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 ?
 
  • #6
GrandMaster87 said:
Wont cosx be Cos0 ?
Exactly. And cos0 equals what?
 
  • #7
Doc Al said:
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
 

1. What is the work energy theorem?

The work energy theorem is a fundamental principle in physics that relates the work done on an object to its change in kinetic energy. It states that the net work done on an object is equal to the change in its kinetic energy.

2. How is the work energy theorem represented mathematically?

The work energy theorem can be written as W = ΔKE, where W is the net work done on the object and ΔKE is the change in kinetic energy. It can also be written in terms of force and displacement as W = Fd cosθ, where F is the applied force, d is the displacement, and θ is the angle between the force and displacement vectors.

3. What are the units of work and energy?

The units of work and energy are the same, and are typically measured in joules (J) in the SI system. However, they can also be expressed in other units such as foot-pounds (ft-lb) or calories (cal).

4. How does the work energy theorem apply to real-world situations?

The work energy theorem can be applied to various real-world situations, such as calculating the work done by a force to lift an object, the work done by a person walking up a flight of stairs, or the work done by a car's engine to accelerate. It is also used in fields like engineering and mechanics to understand and design systems that involve work and energy.

5. What are some limitations of the work energy theorem?

The work energy theorem assumes that there are no other external forces acting on the object besides the applied force. It also assumes that the object is moving in a straight line and that there is no change in potential energy. In reality, these conditions may not always be met, and the work energy theorem may not accurately describe the situation.

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