# Not sure how to plug in numbers for Work Energy Theorem

• CiCi
In summary: Well, the title of the thread mentions use of the work-energy theorem which implies calculation of the sum of all the works. None of the acting forces (gravity, friction and normal force) are in the direction of the displacement.
CiCi
1. The Problem Stament, all variables and given data

a 15 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20 degrees with the horizontal. if there is a constant force of kinetic friction of 25 N between the crate and ramp, what kinetic energy would the crate have at the bottom of the ramp?

2. Needed equations

Work = PE + KE + Energy Lost
PE= mgh
Height = Sinθ * Length
KE= (1/2) m*v^2

3. Attempt at solution

okay so I know how to start out with finding PE

PE= mass* gravity* height = 15 * 9.8 * height

Find Height
Height= Sin (20°) * 2.0= .684

PE= 15 * 9.8 * .684 = 100.5 Joules

What's throwing me off is the kinetic friction of 25 N, I'm not sure what to do with it and where to plug it into to

How can you include the work done by the friction into your energy considerations?

Orodruin said:
How can you include the work done by the friction into your energy considerations?
Is there another equation I need? Because I'm not sure

What equation relates force and distance to work?

CWatters said:
What equation relates force and distance to work?
Force = distance / time ?

I said force, distance and work.
(Not time)

Bystander
CWatters said:
I said force, distance and work.
(Not time)
Work = force * distance ?

CWatters
That's the one.

CWatters said:
That's the one.
So
Work = 25 N * 2.0 m
= 50 Joules
What does this do?

That describes the amount of energy lost to friction. You know that this is the only way energy is lost during the motion of the object. Therefore you can say that the final energy of the box will be the initial energy minus the energy lost to friction.

CWatters
+1

You have a system where energy changes from PE to KE and heat. Apply the relevant equation you posted that invokes conservation of energy.

CiCi said:
Work = force * distance ?
It's more than that.
For a constant force, the work is given by ##W=Fd\cos \theta##, where ##F## = magnitude of the force, ##d## = magnitude of the displacement and ##\theta## = angle between force vector and displacement vector. Work is "force times distance" only if the force and the displacement are in the same direction. This is not the case here.

CWatters
kuruman said:
It's more than that.
For a constant force, the work is given by ##W=Fd\cos \theta##, where ##F## = magnitude of the force, ##d## = magnitude of the displacement and ##\theta## = angle between force vector and displacement vector. Work is "force times distance" only if the force and the displacement are in the same direction. This is not the case here.
That's correct but I hope it doesn't confuse cici. In this case the force and displacement arent in the same direction but are parallel so trig isn't really needed to solve.

CWatters said:
That's correct but I hope it doesn't confuse cici. In this case the force and displacement arent in the same direction but are parallel so trig isn't really needed to solve.
Well, the title of the thread mentions use of the work-energy theorem which implies calculation of the sum of all the works. None of the acting forces (gravity, friction and normal force) are in the direction of the displacement. I hope @CiCi does not get confused either, but he has to understand how the more general definition of work is implemented.

CWatters

## What is the Work Energy Theorem?

The Work Energy Theorem is a fundamental principle in physics that states that the net work done on an object is equal to the change in its kinetic energy.

## How do I use the Work Energy Theorem?

To use the Work Energy Theorem, you must first identify the initial and final states of the object, and then calculate the work done on the object by all external forces. Finally, you can use the theorem to find the change in the object's kinetic energy.

## What are the units for the Work Energy Theorem?

The units for the Work Energy Theorem are joules (J), which is the unit for both work and energy. This is because the theorem equates the work done on an object to the change in its kinetic energy, which is a form of energy.

## Can the Work Energy Theorem be used for all types of motion?

Yes, the Work Energy Theorem can be used for all types of motion, including linear, rotational, and even combined motion. It is a general principle that applies to all types of objects and their motion.

## Are there any limitations to the Work Energy Theorem?

The Work Energy Theorem only applies to conservative forces, which are forces that do not dissipate energy. Additionally, it does not take into account other forms of energy, such as potential energy, and therefore, may not give a complete picture of the system's dynamics.

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