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Two Physics problems.. Work and Electricity! 
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#1
Jan2505, 01:44 PM

P: 390

Here're three problems I have trouble solving, can someone please point out what formula I should be using?
1) An adult pulls a child across the ice on a sled. The combined weight of the sled and the child is 26N, and the adult pulls on a rope that joins the sled at an angle of 30deg (Pi/6 Rad) above the horizontal. Ignoring friction, how much work is done in pulling the sled 390m? I did: 26 = F sin 30deg F = 52N Now, should I do W = Fd or W = Fdcos30? And why? 2) There are three charges, A, B, and C. A and B are vertically seperated by a distance of 0.060m and A has the charge 4.2 x 10^5C and B has the charge 8.4x10^5C. B and C are horizontally seperated by a distance of 0.140m, and C has a charge of 2.1x10^5C. Now, when you're calculating the total charge on B, you're finding the resultant vector.. What if this question said "Find the total charge (force) acting on charge A" or C? Is that even possible? If so, how would that be calculated.. 3) An engineer can vary the rate at which water falls directly down through a generator at a hydroelectric power plant. The generator and its turbine can supply 268MW of electric power when the rate at which the water supplies power to the turbine is 335MW. a) Efficiency? Overly simple, 268/335 = 80% b) What is the change in potential energy of the falling water per second? No clue.. :\ c) If water falls 16m to the turbine, what mass of water must be directed through the turbine per second to supply the power indicated above? Still.. :\ I hope someone can help out with these questions.. I'd appreciate it as my finals are tomorrow. Thanks. 


#2
Jan2505, 02:15 PM

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Though the number involved in problem #1 is pretty hard to believe,i'd say that the problem is not solvable as it is incomplete...
Post your work for problems #2 & #3.For the second,you simply need to compute the electric force...For the 3rd you must know the connection between power & energy. Daniel. 


#3
Jan2505, 02:25 PM

P: 55

I agree that the numbers are unrealistic but why is it unsolvable ?
Work is simply [tex] W = F . d[/tex] Pseudo: You need to multiply it by [tex]cos\theta[/tex] because [tex]sin\theta[/tex] would give you the vertical component of the force. You need the horizontal component. *baa i m confused now.* 


#4
Jan2505, 02:26 PM

P: 390

Two Physics problems.. Work and Electricity!
Well, I don't really understand how to do that, I mean... how do I find the force on A or C but not B? It's kinda like this:
A    BC I only know how to find the resultant force for b, but how about A or C? I don't have any working for this one, as it was a former test question... For 3, I still don't understand how to do any of it. Any ideas? Thanks. 


#5
Jan2505, 02:29 PM

P: 55

For 3b use [tex]P=\frac{W}{t}[/tex]



#6
Jan2505, 02:39 PM

P: 390

Oh, OK.. I can't believe I was blind to number 3 like that.. :P
But number 1 didn't work with cos X, so I wasn't sure whether or not it was right... How about number 2? 


#7
Jan2505, 03:28 PM

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Okay,guys,apparently we're not on the same wavelength here.I ask you both:at the first problem,what is tha force that the adult exerts on the rope (implicitely on the sled+child)...?
Daniel. 


#8
Jan2505, 03:46 PM

P: 55

I was also thinking of the same thing but then i remembered similar problems such as:
+What is the change in the potential energy of an object if we lift it by 10m ? And for this we would use [tex]P_E=mgh[/tex] and [tex]P_E=\Delta W[/tex] How is this different from his question ? 


#9
Jan2505, 03:48 PM

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I think we're not speaking about the same thing.I was referring to the first problem.Your formulas would deal with the third one...
Daniel. 


#10
Jan2505, 03:57 PM

P: 55

OK fine. I was looking at the question from a different perspective.
And saying "we r not on the same wavelength" is so like you. I wouldnt expect it from anyone else. *lol* Well then, then the [tex]\Delta W[/tex] for question 1 should be equal to 0. Since we r assuming the ground is frictionless, any permament force applied would cause it to accelerate and it wouldnt stop. An instantenous force would make it move at constant velocity forever so you need to apply an opposite force to stop it. In the end, lets say [tex]\Delta W=0[/tex] and be happy with it since there is no real solution to it ? :) 


#11
Jan2505, 03:57 PM

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You can't do number 1 because you don't know the force the man is applying to the sled. Without friction, the mass is irrelevant any force would cause an acceleration. If there were friction, then the force would have to be at least the friction force.



#12
Jan2505, 04:28 PM

P: 390

Is the same total force applied on B = total force on C = total force on A in number 2?



#13
Jan2505, 04:47 PM

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No,the forces should be different,since the distances and the magnitudes and the directions are different...Ad those vectors & convince yourself...
Daniel. 


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