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Trigonometry Trigonometry in mechanics

Needhelp

New member
Apr 9, 2012
17
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
 

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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
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?
 

Needhelp

New member
Apr 9, 2012
17
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

Indicium Physicus
Staff member
Jan 26, 2012
4,197
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.