How do I accurately calculate work done against friction?

[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
answer in kilo-joules (kJ)

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

 

[haruspex]

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.

 

[Elara04]

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

 

[haruspex]

10500 is force in newtons instead of kilonewtons

It says 10.1kN

8.9m

It says 9m

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?

 

[Chestermiller]

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)?

 

[nasu]

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.

 

[kuruman]

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.

 

[haruspex]

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

Please try to answer my questions in post #4.
What forces act on the object?
Which of them is f in your w = fs + mgh in post #1?

 

[String theory guy]

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?

 

[haruspex]

Only true for constant velocity?

No, it doesn’t depend on time in any way.

 

[String theory guy]

No, it doesn’t depend on time in any way.

Sorry, why?

 

[haruspex]

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.