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Finding force from work 
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#1
Oct1705, 10:14 AM

P: 43

A 1200kg car is being driven up a 7.76° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 496 N. A force F is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight W and the normal force FN directed perpendicular to the road surface. The length of the road up the hill is 204 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 197 kJ?
in doing this problem i got the work of the force F to be 204*Force of F, the work of the gravity to be 3.24E5, the work of the normal force to be 0, and the work of the kinetic friction to be 1.01E5. The problem says that the net work is 197000 J so i added all those forces and made them equal to that and then i solved for F, but i got a negative force (1117.65), does that make sense? 


#2
Oct1705, 10:19 AM

HW Helper
P: 397

It would be helpful to see where some of those numbers come from. For instance, I'm getting a very different answer for the "work due to gravity". Also, it's useful in cases like this to distinguish between forces that do positive work on your system and those that do negative work  i.e. those that add energy and those that take it away. Which is which here?



#3
Oct1705, 10:29 AM

P: 43

the work due to gravity was obtained by (mg)*(d)*(cos97.76) = (1200)(9.8)(204)(cos97.76) = 3.24E5. negative work i think is gravity and kinetic friction. did i find the work due to gravity the right way?



#4
Oct1705, 10:33 AM

HW Helper
P: 397

Finding force from work
Not really. Tell me  in what form is the "work done by gravity" going to show up? And what is the standard formula for that? Sketch the problem out and I think a little trigonometry will show you what you're doing wrong. Remember  cosine is adjacent over hypotenuse, sine is opposite over hypotenuse.



#5
Oct1705, 10:48 AM

P: 43

"work done by gravity" should be negative, right? the standard formula for that is Wg=Fg*d*cos(theta), the Fg is mg and since the mass is 1200 kg and g is 9.8 wouldn't Fg be 11760? the distance is 204, and theta would be 90 degrees + 7.76 degrees = 97.76 degrees (wouldn't it be 97.76 because the angle is clockwise?) because the weight is pointing straight down, not perpendicular to the ramp. i drew it out and thats what i got and i'm not really sure whats wrong there.



#6
Oct1705, 10:57 AM

HW Helper
P: 397

Short lecture time: There's a strong tendency among beginning physics students to treat the subject as just a great ocean of formulas from which you have to fish the right one to solve a problem. This tendency should be resisted. For one thing, it's quite possible to remember the formulas incorrectly, leading one to possibly interesting but physically impossible solutions. You have to look at each problem with a mind to what's happening physically if you ever hope to understand the relationship between physical reality and the mathematical models you're building in class.
When you're dealing with gravity, you're pretty much automatically dealing with the vertical, right? The work done against gravity is going to show up as a change in the gravitational potential energy of the object  do work against gravity and you're going to end up with an increase in the gravitational potential energy, which will translate into an increase in the height of the object. Allow gravity to do the work and the object will lose PE (which will, of course, appear somewhere else) with a corresponding decrease in height. Given that it's the height of the object that matters, go back and look at your diagram again. When you're dealing with the length of the hill, you're dealing with the hypotenuse of a right triangle. Which leg do you want if you're looking for a change in height? And which of the trig functions deals with that leg and the hypotenuse? Also  an angle of 97 degrees is going to give you an obtuse triangle. Your diagram, on the other hand, probably involves a right triangle, which pretty much rules out obtuse. Don't make the problem more difficult than it really is. Does that help? 


#7
Oct1705, 11:13 AM

P: 43

so when i figure out the work of gravity i multiply the force of gravity times the distance (which would be according to the height (27.54) found by caculating 204 m * sin(7.76) )... i'm still confused because the weight points straight down, not parallel to how i labelled my yaxis (its 7.76 degrees left of my yaxis), so how do i put that into my equation?



#8
Oct1705, 11:55 AM

HW Helper
P: 397

You have to distinguish between vector quantities, where direction matters, and nonvector quantities where it doesn't. In this particular case, the change in potential energy is a scalar  it doesn't matter how you orient your axes, only where you've put your reference plane. Your reference plane is the bottom of the hill, and the car ends up 27.5 m above that, so you have a particular change in its potential energy which is independent of how you orient your axes.
Does that make sense? 


#9
Oct1705, 01:06 PM

P: 43

yes but then when does the cos(theta) of the equation W=F*d*cos(theta) come into play?



#10
Oct1705, 01:25 PM

HW Helper
P: 397

It would depend on the circumstances, luv. I would guess that's the work done by a force pushing something (for instance) across the floor at an angle of theta with the horizontal. In a case like that, only the horizontal component of the force does any work  hence cosine.



#11
Oct1705, 01:37 PM

P: 43

oh okay thanks



#12
Oct1705, 05:06 PM

P: 43

i'm still not getting this problem... i found that the work of kinetic friction is the work of F by using the h (since i found it from the distance) and then i broke the Work of gravity into x and y components and found the x component of the work of gravity. i then added all of the works together to equal the net work and then i solved for F but i'm still not getting the right answer.. help?!



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