Why Is No Work Done Against the Weight of a Skier Moving Down a Slope?

[Needhelp2]

Hi!
I am finding the work-energy principle and idea of total mechanical energy hard to apply to finding the work done against/by a particular force.
For example in the question below, why is there no work done against the weight of the skier acting down the slope?

Any help would be great!
Thanks :D

 

[Ackbach]

So don't forget the sign conventions with work. Since work done by a constant force is actually defined by $W= \vec{F} \cdot \vec{d}=F\,d\, \cos(\theta),$ where $\theta$ is the angle between the force vector $\vec{F}$ and the displacement vector $\vec{d}$, what you are asked to do in this problem is combine two different expressions for work (definition and the work-energy theorem) to find the work done by the pulling force.

I would recommend a free-body diagram for this problem. One thing you know: the increase in speed from $2$ m/s to $5$ m/s gives you a change in kinetic energy, which is equal to the net work. The net work comes from the net force. If you can find the net force, you could probably find the pulling force, so long as you know all the other forces. Then what could you do?

 

[Needhelp2]

So don't forget the sign conventions with work. Since work done by a constant force is actually defined by $W= \vec{F} \cdot \vec{d}=F\,d\, \cos(\theta),$ where $\theta$ is the angle between the force vector $\vec{F}$ and the displacement vector $\vec{d}$, what you are asked to do in this problem is combine two different expressions for work (definition and the work-energy theorem) to find the work done by the pulling force.

I would recommend a free-body diagram for this problem. One thing you know: the increase in speed from $2$ m/s to $5$ m/s gives you a change in kinetic energy, which is equal to the net work. The net work comes from the net force. If you can find the net force, you could probably find the pulling force, so long as you know all the other forces. Then what could you do?

you haven't mentioned the gain in GPE the skier would have or the work done against the friction? Where would these come in?

 

[Ackbach]

you haven't mentioned the gain in GPE the skier would have

Well, the problem here is that energy is not conserved, since there is friction. Therefore, a conservation of energy approach is invalid. Hence, you must analyze the forces vectorially. Gravity definitely plays a role, but it'll show up in your force analysis, rather than as a gravitational potential energy.

or the work done against the friction?

There are three forces doing work (that is, there are three forces that are either parallel or anti-parallel to the displacement): the pulling force (doing positive work), the friction force (doing negative work), and the component of gravity that is directed down the incline (doing negative work). Those all have to show up when you write down the net work.

Where would these come in?

Hopefully, I've answered this question. But by all means, if you're still stuck, keep pushing me with more questions.

 

[CellCoree]

Hi there,

Trigonometry is a fundamental mathematical tool used in mechanics to understand and analyze the motion and forces acting on objects. It allows us to break down complex forces into their components and calculate their magnitudes and directions.

In the case of the work-energy principle and total mechanical energy, trigonometry is essential in determining the work done by or against a particular force. The work done by a force is equal to the magnitude of the force multiplied by the displacement in the direction of the force. Trigonometry helps us to determine the displacement in the direction of the force, as well as the angle between the force and the displacement.

In the question you mentioned, the skier is moving down a slope, which means the direction of motion and the direction of the weight force (acting down) are the same. In this case, the angle between the force and the displacement is 0, and therefore, the work done by the weight force is also 0. This is because the work done by a force is only non-zero when there is a component of the displacement in the direction of the force.

I hope this explanation helps you understand the role of trigonometry in mechanics and how it can be applied to solve problems involving work and energy. Keep practicing and don't hesitate to ask for help when needed.

Best of luck!