Kinetic energy- pushing an object up an incline.

  • Thread starter bopll
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  • #1
bopll
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Homework Statement


An object is given an initial velocity of 3.0 m/s up a 20 degree incline. Mass= 22.3 kg and coefficient of kinetic friction is .93.


Homework Equations



W = (1/2)mv1^2-1/2mv0^2

W = Fdcos(u)


The Attempt at a Solution



So I added the forces together, mgsin(20) + .93mgcos(20)

and this times cos(20) should be the total work, right?

so then since the change in kinetic energy equals work i set -(1/2)mv^2= -Fdcos(20)

plugging in the mass for m, initial velocity for v, the forces due to friction and gravity for F, and solving for d, i got the wrong answer. where'd i go wrong? I also tried this taking out the cos(20) in the second work equation and still got it wrong.


Thanks, this is my first time using physics forums and i really hope i can get to understand this stuff...
 

Answers and Replies

  • #2
Rake-MC
325
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What is the question asking for?
 
  • #3
bopll
13
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sorry. the distance it's pushed up the hill.
 
  • #4
Rake-MC
325
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Well I say you've done it fine by adding up the two forces.
What I would have done then would be to tilt the axes 20 degrees such that it's parallel with the incline. We'll call this axis the x.

We know that the magnitude of net force acting on the x axis is mgsin(20) + 0.93cos(20).

Therefore just use [tex] \frac{1}{2}mv^2 = F\Delta d [/tex]
 
  • #5
bopll
13
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Well I say you've done it fine by adding up the two forces.
What I would have done then would be to tilt the axes 20 degrees such that it's parallel with the incline. We'll call this axis the x.

We know that the magnitude of net force acting on the x axis is mgsin(20) + 0.93cos(20).

Therefore just use [tex] \frac{1}{2}mv^2 = F\Delta d [/tex]


that's what i did :-\

it came out to .3979 ish and that wasn't right...
 
  • #6
bopll
13
0
ugh. my calculator was in radians.

*facepalm*
 
  • #7
bopll
13
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How do i mark a thread as solved?
 
  • #8
Rake-MC
325
0
You can just let it die away
 

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