AAMC Assessment - Physics

[dnovikov]

---

Members don't see this ad.

A little confused about how AAMC went about answering this question:

Answer choices:

So I thought that Work is calculated using the Force thats parallel to displacement. If thats so I assumed you needed to multiply the Force vector by its component in the same direction as the movement of the weight? Any help in clearing this up is much appreciated! Maybe I am just overthinking it...

 

[Jepstein30]

A little confused about how AAMC went about answering this question:

Answer choices:

So I thought that Work is calculated using the Force thats parallel to displacement. If thats so I assumed you needed to multiply the Force vector by its component in the same direction as the movement of the weight? Any help in clearing this up is much appreciated! Maybe I am just overthinking it...

Machines don't change the amount of work done. So we need to raise a 4 kg mass by 5 m.

W = mgh = (4)(5)(10) = 200 J

That's how I approached that problem. Even thinking about the machine is the wrong way to go. To answer your question though, in this case, the tension in the rope is what is actually doing the work and THAT is parallel to the displacement. The force is providing the tension.

Overthinking it though! If you ever see work mentioned with machines, its usually a trick question!

 

[dnovikov]

Machines don't change the amount of work done. So we need to raise a 4 kg mass by 5 m.

W = mgh = (4)(5)(10) = 200 J

That's how I approached that problem. Even thinking about the machine is the wrong way to go. To answer your question though, in this case, the tension in the rope is what is actually doing the work and THAT is parallel to the displacement. The force is providing the tension.

Overthinking it though! If you ever see work mentioned with machines, its usually a trick question!

I see, thanks for the quick response!

 

[ozzi22]

so why wouldn't it be Fdcos theta making the answer 100? Isn't that the equation for force?

 

[indianjatt]

so why wouldn't it be Fdcos theta making the answer 100? Isn't that the equation for force?

you could be one of two ways. One is to notice that work in has to equal work out. You must conserve energy or find out the vertical component of the force which will be used move the block, which is sin30*40N=20N. Work = Fd. Since you are using a mechanical machine with an advantage of two, the force is one half.... But the distance is 2d. W = 20*10= 200 J

 

[ozzi22]

you could be one of two ways. One is to notice that work in has to equal work out. You must conserve energy or find out the vertical component of the force which will be used move the block, which is sin30*40N=20N. Work = Fd. Since you are using a mechanical machine with an advantage of two, the force is one half.... But the distance is 2d. W = 20*10= 200 J

Thanks 👍