Babynini
  • Babynini
Force and work.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Babynini
  • Babynini
A car drives at 800ft. on a road inclined at 9 degrees. Car weighs 2700lb. Thus gravity acts straight down on the car with a constant force of f=-2700j. Find the work done by the car in overcoming gravity.
Babynini
  • Babynini
Um..what i've done so far is |v|=2700cos(9) = 2666.76 which is the force experienced by the driveway. Is that the answer i'm looking for?
Babynini
  • Babynini
@ganeshie8

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Babynini
  • Babynini
@zepdrix :)
Babynini
  • Babynini
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
Work = Force * Displacement
jim_thompson5910
  • jim_thompson5910
Gravity is pulling straight down, so when we think of "displacement" we are only considering the vertical displacement of the car
jim_thompson5910
  • jim_thompson5910
|dw:1433399821064:dw|
jim_thompson5910
  • jim_thompson5910
when you wrote `Thus gravity acts straight down on the car with a constant force of f=-2700j` that doesn't make any sense. Joules aren't a measure of force. They are measure of energy. You might be thinking of newtons?
Babynini
  • Babynini
uhh idk that's what my homework says.
jim_thompson5910
  • jim_thompson5910
oh nvm, -2700j is a vector, not 2700 joules
Babynini
  • Babynini
haha yeah
Babynini
  • Babynini
where is hte 800 from?
jim_thompson5910
  • jim_thompson5910
"A car drives at 800ft"
jim_thompson5910
  • jim_thompson5910
I'm assuming it drives up that incline and drives 800 ft
Babynini
  • Babynini
ah yeah k.
jim_thompson5910
  • jim_thompson5910
use trig to find the value of y in the drawing below |dw:1433400113262:dw|
Babynini
  • Babynini
125.148 ?
jim_thompson5910
  • jim_thompson5910
so the car essentially goes up vertically 125.148 ft this is the vertical displacement we care about
Babynini
  • Babynini
(ps: it says you are expected to use vectors and the appropriate formulas to find work. Just as a note. :P)
Babynini
  • Babynini
oo ok
Babynini
  • Babynini
so now we have 2=-2700 * 125.148
Babynini
  • Babynini
that 2 was meant to be a w
Babynini
  • Babynini
er. that will make a very nasty negative number..
jim_thompson5910
  • jim_thompson5910
were you given the vector formula for work in physics?
Babynini
  • Babynini
This is precalc.
Babynini
  • Babynini
give me a moment to find it, I have a formula somewhere :)
Babynini
  • Babynini
Yep w=f*d
jim_thompson5910
  • jim_thompson5910
I'm thinking you basically take the force needed to climb that vertical distance and multiply it with the vertical distance I'm not sure how to fit in vector formulas though
Babynini
  • Babynini
so an example given is: A man pulls a wagon horizontally by exerting a force of 20 lb on the handle. If the handle makes an angle of 60 degrees with horizontal, find the work done in moving the wagon 100ft Solution: We choose a coordinate system with the origin at the initial position of the wagon. That is, the wagon moves from point p(0,0) to the point Q(100,0) the vector that represents this displacement is d=100i The force on the handle can be writen in terms of components as F= (20cos(60))i + (20sin(60))j F = 10i+10 sq3j Thus the work done is (10i+10 sq3j)*(100i)=1000 = 1000
Babynini
  • Babynini
so our displacement is D = 800i
Babynini
  • Babynini
so it's written as F=(2700cos9)i+(2700sin9)j = 2666.785i+422.373j
Babynini
  • Babynini
and work done is = (2666.785i+422.373j)*(800i)
Babynini
  • Babynini
I'm not sure how they calculated the last part though.
Babynini
  • Babynini
Is this kind of making sense? I'm following formulas mostly. o.0
Babynini
  • Babynini
@jim_thompson5910 :P
Babynini
  • Babynini
(sorry, i'll be patient ;))
jim_thompson5910
  • jim_thompson5910
sorry I got distracted for a sec but it looks like you're on the right track. You dot product the force vector F and the displacement vector d to get the work
Babynini
  • Babynini
it's ok! I know you're helping other people. Keep up the hard work :)
Babynini
  • Babynini
so is 2133829.189 the final answer you're getting?
jim_thompson5910
  • jim_thompson5910
The force of gravity is F = 0i + (-2700j) essentially this vector pulls the car straight down with magnitude 2700 lb -------------------- The displacement vector d = 800*cos(9)i + 800*sin(9)*j -------------------- dot product the two vectors W = F dot d W = [0i + (-2700j)] dot [800*cos(9)i + 800*sin(9)*j] W = [0*800*cos(9)] + [-2700*800*sin(9)] W = -337,898.444486899 That seems like a pretty massive number, so I'm not 100% sure
Babynini
  • Babynini
bleh yeah that's pretty huge :/
Babynini
  • Babynini
wait why 800 instead of 2700 before the cos and sin?
jim_thompson5910
  • jim_thompson5910
because if we ignore the weight, and we just consider moving along the hypotenuse, we basically have this triangle |dw:1433402120340:dw|
Babynini
  • Babynini
but in the example one they used the weight of the cart in that spot of the equation, which is why I ask. hm but yeah I see what you're saying.
jim_thompson5910
  • jim_thompson5910
|dw:1433402181444:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1433402210524:dw|
jim_thompson5910
  • jim_thompson5910
hmm now I'm not sure. I've looked back at your example and you're right
Babynini
  • Babynini
lol. What is this madness.
anonymous
  • anonymous
\[W=\vec F. \vec s=Fs \cos(\theta)\]
anonymous
  • anonymous
Everything is given, just convert to standard units and use formula, feet to metres, pounds to kilograms and degrees to radians
anonymous
  • anonymous
|dw:1433411117463:dw|
Babynini
  • Babynini
@rational how do I solve this? There's so many formulas that have been given here D:
Babynini
  • Babynini
F = -2700j so = <0,-2700> Displacement (or shift) = 800 feet. so = <800,0> all correct so far?
jim_thompson5910
  • jim_thompson5910
<800,0> means you go 800 ft to the right and 0 ft up/down
jim_thompson5910
  • jim_thompson5910
<800,0> = 800i + 0j
Babynini
  • Babynini
Yep yep. So now according to W = F * D I dot product <0,-2700> * <800,0> ?
jim_thompson5910
  • jim_thompson5910
but the object isn't going on a flat level plane its on an inclined plane
Babynini
  • Babynini
oh yeh.
Babynini
  • Babynini
where is that incorporated?
jim_thompson5910
  • jim_thompson5910
think of the point \(\Large (r, \theta) = (800, 9^{\circ})\) what is the rectangular (x,y) form of this point?
Babynini
  • Babynini
what?
jim_thompson5910
  • jim_thompson5910
|dw:1433463727512:dw| what are x and y?
Babynini
  • Babynini
Opposite and Adjacent? is that what you're asking?
jim_thompson5910
  • jim_thompson5910
Once you find x and y, you will be able to represent the displacement as a vector
jim_thompson5910
  • jim_thompson5910
you use trig to find x and y sin(theta) = opp/hyp sin(9) = y/800 y = 800*sin(9) do the same with x (but use cosine instead)
Babynini
  • Babynini
y= 125.149 x=790.151
jim_thompson5910
  • jim_thompson5910
|dw:1433464222924:dw|
jim_thompson5910
  • jim_thompson5910
so the displacement vector is <790.151, 125.149> or 790.151i + 125.149j
Babynini
  • Babynini
and Force still is = <0,-2700> ?
jim_thompson5910
  • jim_thompson5910
I'm realizing now that's where the mistake is that's the force of gravity, but we want the force of the car if it asked "what is the work done by gravity?" then we would use F = <0,-2700>
jim_thompson5910
  • jim_thompson5910
but instead it says "Find the work done by the car in overcoming gravity."
Babynini
  • Babynini
aah hrm. so something to do with the weight?
Babynini
  • Babynini
Wait, but it says "the gracity acts straight down on the car with a constant force of -2700j." the force pushing against the car would be the same the car is pushing back though. sot hat would be the force the gar is "pushing against gravity"?
Babynini
  • Babynini
*gravity
jim_thompson5910
  • jim_thompson5910
maybe to overcome the force of -2700j, the car needs to apply 2700 lb somehow?
jim_thompson5910
  • jim_thompson5910
yeah as you just said
jim_thompson5910
  • jim_thompson5910
I'm not sure though
Babynini
  • Babynini
But -2700 is what is being pushed against the car. Which is the same amount the car is pushing back against gravity. So that should still be the force.
jim_thompson5910
  • jim_thompson5910
that might be correct
Babynini
  • Babynini
...
jim_thompson5910
  • jim_thompson5910
I found this page http://cposcience.com/home/Portals/2/Media/post_sale_content/PHY2/Ancillaries/SkillSheets/Unit_4/10.2_WorkDoneagainst.pdf
jim_thompson5910
  • jim_thompson5910
sorry I'm not that great at physics
Babynini
  • Babynini
apparently neither am I ;P
jim_thompson5910
  • jim_thompson5910
if you use that link's formula, be careful to make sure things are converted to metric
Babynini
  • Babynini
er but this is vector stuff not joules and what not
jim_thompson5910
  • jim_thompson5910
pounds ---> kilograms feet ----> meters
jim_thompson5910
  • jim_thompson5910
joules is just the units of work in physics
Babynini
  • Babynini
Ah k.

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