aamc physics question on pulleys that goes contrary to what TBR says. help!!

[wellcutcookies]

A force F is used to raise a 4-kg mass M from the ground to a height of 5 m

What is the work done by the force F? (Note: sin 60o = 0.87; cos 60o = 0.50. Ignore friction and the weights of the pulleys.)

50 J
100 J
174 J
200 J

Answer: D

Explanation: Work is the product of force and distance. The easiest way to calculate the work in this pulley problem is to multiply the net force on the weight mg by the distance it is raised: 4 kg x 10 m/s2 x 5 m = 200 J. Therefore, answer choice D is the best answer.

I thought the answer would be 100 J, seeing as 2F = mg, so F=mg/2 = 20 N. 20 N times 5 m = 100 J.
i've checked my TBR book (pg 97 of the first physics book if anyone else has it), and they say the force would be halved in this scenario (they have an identical scenario in their book).... I agree with TBR, as that's the whole point of a pulley. the aamc, however, say that the force F provided is 40 N, which would be true if there were no pulley at all, but instead just a single rope holding up the block. i feel like the aamc is wrong here. can anyone explain??

thanks

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[kevinus]

Work is the integral of force vector dot movement vector. The mass only moves downward, so the portion of the force vector that applies to the work done on the mass is mg. F*d=mgd=4*10*5=200J

 

[pjk77]

the point of a pulley or any machine is to decrease the force by increasing the distance. overall work (F x d) is the same.

 

[Cmdr_Shepard]

The two cables pulling up oppose the 40 N of the mass. Each cable takes half the weight so each cable takes a load of 20 N.

Then, since the tension in the cable is the same throughout the entire length, the value of F is 20 N. However, if the weight is raised by 5 m and it weighs 40 N, then you need to input 200 J of work. This can only happen if the rope is pulled 10 metres since the tension is only 20 N.

 

[dudewheresmymd]

Apply energy conservation--> no friction--> only conservative forces acting --> Wc=-delta PE --> mgh --> 4*10*5 --> 200 J, here 40N across 5m, (LARGER force, SHORTER DISTANCE without a pulley)

Using TBR approach, T =mg/2 --> tension in rope is 20 N, Work has to stay same therefore:

200J = F*d
200= (20N) *10m (SMALLER force, LONGER distance when using a pulley)

Now, you have to apply 20N across 10m to get the same amount of work. You incorrectly assumed that the 20N would be applied over the same distance, however, pulleys do just the opposite. They reduce force needed by increasing the distance the force is applied upon.

Bonus: What is the mechanical advantage of the pulley system above?

Weight of object / Applied force needed to support object= (40N/20N) = 2 (same as # of vertical ropes that support the hanging weight 2 for mass M above)

Hope this helps.

 

[wellcutcookies]

thanks guys!! i appreciate the clarification. it seems i had a very fundamental misunderstanding, which is really bad

 

[-----MD]

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[ilovemedi]

nvm....

 

[whatwhy]

Why wouldn't this be f*d*cos60? Instead of just f*d? I got 100, not 200.

 

[theonlytycrane]

One more quick question: The two cables pulling up oppose the 40 N of the mass. Each cable takes half the weight so each cable takes a load of 20 N.

Doesn't this only balance the mass attached to the pulley? In order to bring it up wouldn't > 20N per cable need to be applied?

(I think I might be confused on the fact that the mass can move up without a net force up)

 

[amy_k]

ignoring friction and any inefficiencies of the pulley, only at least 20N will be needed to raise the mass. If it were already in motion, a constant exerted force of 20N would cause it to continue moving upwards at constant speed (because forces are balanced so acceleration is zero). Usually in real life a slightly greater force is required initially to set a static object in motion, due to static friction. But since we ignore friction here it is just 20N

 

[Quinoline]

Why wouldn't this be f*d*cos60? Instead of just f*d? I got 100, not 200.

W=F*d*cos(angle between F and d)
In this case, the force acting on the object is directly up, and the displacement is up, so the angle is 0.

 

[theonlytycrane]

ignoring friction and any inefficiencies of the pulley, only at least 20N will be needed to raise the mass. If it were already in motion, a constant exerted force of 20N would cause it to continue moving upwards at constant speed (because forces are balanced so acceleration is zero). Usually in real life a slightly greater force is required initially to set a static object in motion, due to static friction. But since we ignore friction here it is just 20N

Thanks, amy_k! One more clarification- so if we ignore friction, 20N per cable will either balance the mass (no movement) or keep the mass moving upwards at a constant speed (assuming it was already in motion?). The main point is that it won't be accelerating. Am I thinking about this correctly?

 

[sazerac]

Yes

 

[amy_k]

Thanks, amy_k! One more clarification- so if we ignore friction, 20N per cable will either balance the mass (no movement) or keep the mass moving upwards at a constant speed (assuming it was already in motion?). The main point is that it won't be accelerating. Am I thinking about this correctly?

yes you got it

 

[whatwhy]

W=F*d*cos(angle between F and d)
In this case, the force acting on the object is directly up, and the displacement is up, so the angle is 0.

But dont they literally show that F is at an angle? I feel so stupid...

 

[Quinoline]

But dont they literally show that F is at an angle? I feel so stupid...

The rope is pulled at that angle, but the force that matters is the force on the object. The rope wraps around the pulley so that the force is actually applied upward on the object. The object doesn't feel what angle you pull the rope at, as long as the rope attached directly to it stays at the same angle.

 

[sazerac]

But dont they literally show that F is at an angle? I feel so stupid...

That angle affects the forces on the upper pulley, which is not relevant to the question they are asking. Theta is a red herring.

 

[catinthehat1234]

W=F*d*cos(angle between F and d)
In this case, the force acting on the object is directly up, and the displacement is up, so the angle is 0.

But isn't the y-component of the force Fsin(angle)? Wouldn't that be the force acting upward along the direction of motion? Sorry to jump in and thanks in advance!

 

[wizzed101]

But isn't the y-component of the force Fsin(angle)? Wouldn't that be the force acting upward along the direction of motion? Sorry to jump in and thanks in advance!

F is applied to counter the torque from the second pulley, not to directly lift the weight.

 

[catinthehat1234]

F is applied to counter the torque from the second pulley, not to directly lift the weight.

I see, thank you. I'm just not sure how to avoid getting into such traps on the MCAT

 

[BerkReviewTeach]

Thanks, amy_k! One more clarification- so if we ignore friction, 20N per cable will either balance the mass (no movement) or keep the mass moving upwards at a constant speed (assuming it was already in motion?). The main point is that it won't be accelerating. Am I thinking about this correctly?

I want to add something here. If we start with the object at rest and it finishes at rest, then at some point it had to be accelerated up (in this case the tension in each pulley cable would have to be greater than 20 N) and it had to be accelerated down (with the tensions being less than 20 causing it to slow and come to rest). The key point here is that the average tension during the lifting process is 20 N for the pulley ropes. So the work done is Favg x d, which for each rope is 20N x 5 m = 100 J per rope, The two ropes combine to do 200 J of work.

As pointed out, it was easier in this question to consider the mass, which has a force down of 40N due to gravity, so it required an average force of 40N up to lift it from rest for 5 meters with it finishing at rest. The work done on the mass is Favg x d = 40N x 5 m = 200 J.

Please disregard the various comments from posters above about friction, because that is NOT entering the question. If there is friction, then you need to do more work than just the change in potential energy. It would equal the change in PE + the work done by friction.

 

[perfect weather]

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