# Work is done against friction

• Elara04
In summary: Because the work done by a force is just ##W=\vec F\cdot\vec s##, where ##\vec s## is the displacement.

#### Elara04

Homework Statement
An object on an inclined plane of weight W=20kN is acted on by a force of 10.1 kN parallel to an inclined plane. The object travels up the slope at a constant speed and travels over a distance s = 37 m, also gaining h = 9 m in height.
How much work is done against friction (i.e. energy dissipated as heat)? Give your
Relevant Equations
𝑊 = 𝐹 ∙ ∆ 𝑠 = 𝐹 ∥ ∆ 𝑠
I'm unsure on where to begin with this question, i've tried many different formulas that aren't giving me the right answer. I believe to start I need to convert the kilo newtons to newtons.
I tried w = fs + mgh
w = 10500 x 8.9/sin(13.9)+(1845.69 x 9.8 x8.9) = 549986.46 J
and then convert to kilo joules = 549.99 kJ

But this isn't correct
any help would be appreciated

Elara04 said:
w = 10500 x 8.9/sin(13.9)+(1845.69 x 9.8 x8.9) = 549986.46 J
Please explain all the parts of that.
Where do 10500, 1845.69 and 8.9 come from?
Why are you dividing by sin(13.9°)?

Your work would be much easier to follow if you were to refrain from plugging in numbers straight away. There are many benefits in working symbolically as far as possible.

MatinSAR
haruspex said:
Please explain all the parts of that.
Where do 10500, 1845.69 and 8.9 come from?
Why are you dividing by sin(13.9°)?

Your work would be much easier to follow if you were to refrain from plugging in numbers straight away. There are many benefits in working symbolically as far as possible.
Sorry, 10500 is force in newtons instead of kilonewtons, 8.9m is the height, 13.9 is theta. I was using w = F x H / sin theta + mgh, although i'm not entirely certain whether that equation is correct

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Elara04 said:
10500 is force in newtons instead of kilonewtons
It says 10.1kN
Elara04 said:
8.9m
It says 9m
Elara04 said:
w = F x H / sin theta
Why H / sin theta when you are given s?
But your big problem there is being clear about what force F is.
What forces act on the object?

MatinSAR
What are the forces acting on the mass in the direction parallel to the slope? Please name them from your free-body diagram. What direction are they pointing (upslope or downslope)?

MatinSAR
The work dissipated by friction has nothing to do with the weight of the body or any other forces. You just need the force, the displacement and the angle between them.

The work done against friction sometimes goes by the name "the work done by friction". So find the force of kinetic friction, find the cosine of the angle between it and the displacement and multiply the three quantities.

Elara04 said:
Hey, so the formula would be w=fd cos theta, but after using that formula I need to find the work done against friction and im not entirely sure how to do that
What forces act on the object?
Which of them is f in your w = fs + mgh in post #1?

nasu
kuruman said:
The work done against friction sometimes goes by the name "the work done by friction". So find the force of kinetic friction, find the cosine of the angle between it and the displacement and multiply the three quantities.
Only true for constant velocity?

String theory guy said:
Only true for constant velocity?
No, it doesn’t depend on time in any way.

String theory guy
haruspex said:
No, it doesn’t depend on time in any way.
Sorry, why?

String theory guy said:
Sorry, why?
Because the work done by a force is just ##W=\vec F\cdot\vec s##, where ##\vec s## is the displacement.
Time, velocity, acceleration do not appear in the formula.

String theory guy