A vector is a magnitude with a direction attached; there's no difference in calculation. Maybe you needed to specify the direction of the gravitational force?
In that case, your work looks right; you just may be missing a force the man has to work against... I suggest what sonic91 said: put everything on a force diagram and go from there.
Your logic looks sound; it must just be a math error throwing you off... If the work you have here is what you typed into your calculator, then I have two suggestions: first, the weight in the 2nd equation should be 427, not 527... secondly, did you take the square root of both sides at the...
Are you sure you meant to say 'normal reaction' twice here? I'm a bit confused by your work; what are you taking the horizontal component of?
Also, what do you mean with fmax? Is that the force the person has to push?
You can find the distance (if you really wanted to) with the definition of electric force, F - (kq1q2)/d2. You know F and the charges, and k's a constant. And distance can't be negative, so you don't have to worry about that when evaluating the square root ;)
(hello classmate! See, I TOLD...
Wait, so you multiplied sin 30 * 5 and got 7.5? Are you sure that you put that into your calculator right? The sine of 30 should come out to be 0.5...
You also need to factor the acceleration due to gravity in there somehow. Or did you already put that in somehow with your 7.5 and just not...
Homework Statement
A hypodermic syringe contains a medicine with the density of water. The barrel of the syringe has a cross-sectional area of 2.5*10^-5 m^2. In the absense of a force on the plunger, making the medicine squirt from the needle. If the syringe is horizontal and the pressure...
I believe you're right about the force, but are you sure you're supposed to take the cosine of 15º? Since Work also equals the change in Potential Energy, it seems like you'd only want the vertical distane the car travels.
(note: not at all to scale)
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y|``\ 150m
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`|___(\<-...
If you've already got the electric field, you don't need a distance; you can just use the fact that F=qE. If your field exerts 2000 Newtons of force on every Coulomb of charge in it, and your particle's charge is -1.6 * 10^-19 Coulombs or whatever (this may not be accurate; I don't exactly know...