# Work Done by Friction & Gravity on Incline: Explained

• simphys
In summary, the work done by gravity is equal to the displacement along the incline multiplied by the cosine of the angle between the incline and the horizontal direction.
simphys
Homework Statement
You are a member of an Alpine Rescue Team. You must project
a box of supplies up an incline of constant slope angle a so that it reaches a
stranded skier who is a vertical distance h above the bottom of the incline.
The incline is slippery, but there is some friction present, with kinetic friction coefficient mk. Use the work–energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it
will reach the skier. Express your answer in terms of g, h, mk, and a
Relevant Equations
##W = Fscos(alpha)##
where ##s=h/sin(alpha)##
So for the work done by the kinetic friction, the displacement along the incline is ##s## as given.
What I canNOT understand is why the displacement in the y-direction is used for the work done by gravity i.e. ##W = -mgh## where ##h## is the displacement in het y-direction. This instead of the displacement along the incline
This is complete bullshit, the work done on a force was supposed to be the component along the displacement wasn't it ?

Edit: this was the answer from the book

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Using the displacement in the force direction and the full force magnitude is completely equivalent to using the force component in the displacement direction and the full displacement.

Orodruin said:
Using the displacement in the force direction and the full force magnitude is completely equivalent to using the force component in the displacement direction and the full displacement.
What a disaster... of course, thanks a lot! it comes down to the same thing but a little bit of a different formula.

The formula is the same, ##W = F d \cos\theta##, the difference is in the interpretation of that formula in words.

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simphys said:
What a disaster... of course, thanks a lot! it comes down to the same thing but a little bit of a different formula.
Our incline of constant slope angle works as a simple machine: the inclined plane.
##W{gravity}~=~-mgh~=~-mgscos\alpha##

Copied from
https://en.wikipedia.org/wiki/Inclined_plane#Inclined_plane_with_friction

"Moving an object up an inclined plane requires less force than lifting it straight up, at a cost of an increase in the distance moved. The mechanical advantage of an inclined plane, the factor by which the force is reduced, is equal to the ratio of the length of the sloped surface to the height it spans. Owing to conservation of energy, the same amount of mechanical energy (work) is required to lift a given object by a given vertical distance, disregarding losses from friction, but the inclined plane allows the same work to be done with a smaller force exerted over a greater distance.
...
The mechanical advantage MA of a simple machine is defined as the ratio of the output force exerted on the load to the input force applied. For the inclined plane the output load force is just the gravitational force of the load object on the plane, its weight Fw. The input force is the force Fi exerted on the object, parallel to the plane, to move it up the plane."

I take it ##\alpha## is supposed to be the angle between the incline and the horizontal direction.

When you write ##W=Fs\cos\alpha##, what does ##\alpha## represent? Is it the same angle as above? If so, are you sure it's the right angle to be using when calculating the work?

Orodruin said:
The formula is the same, ##W = F d \cos\theta##, the difference is in the interpretation of that formula in words.
I prefer the interpretation that treats the vectors on an equal footing, "The work done by a constant force is the product of three quantities: (a) the magnitude of the force vector; (b) the magnitude of the displacement vector; (c) the cosine of the angle between the two."

simphys said:
What a disaster... of course, thanks a lot! it comes down to the same thing but a little bit of a different formula.
In fact for a conservative force (which includes gravity and other forces but not friction) the work done is path independent. The gravitational work calculated along any path from 1 to 2 will produce the same number.
So you can just choose a horizontal segment (Wgrav=0) plus a vertical segment (Wgrav=-mgh) Have you not seen this in your text?

## 1. What is work done by friction and gravity on an incline?

The work done by friction and gravity on an incline is the amount of energy expended or transferred when an object moves along an inclined surface. This work is a result of the force of gravity pulling the object down the incline and the friction between the object and the surface of the incline.

## 2. How is the work done by friction and gravity calculated?

The work done by friction and gravity can be calculated using the formula W = Fd, where W is the work done, F is the force acting on the object, and d is the distance the object moves along the incline. The force, F, is the sum of the force of gravity and the force of friction.

## 3. What factors affect the work done by friction and gravity on an incline?

The work done by friction and gravity on an incline is affected by several factors including the angle of the incline, the mass of the object, the coefficient of friction between the object and the incline, and the distance the object moves along the incline.

## 4. How does the angle of the incline affect the work done by friction and gravity?

The angle of the incline affects the work done by friction and gravity because it determines the amount of force of gravity acting on the object. As the angle of the incline increases, the force of gravity pulling the object down the incline also increases, resulting in a greater amount of work done.

## 5. What is the relationship between work done by friction and gravity and the speed of the object on an incline?

The work done by friction and gravity on an incline is directly proportional to the speed of the object. This means that as the speed of the object increases, the work done by friction and gravity also increases. This is because a faster moving object will cover a greater distance along the incline, resulting in a greater amount of work being done.

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