Work done by gravity on a object on an inclined planeby hyuen9195 Tags: frction, gravity, incline plane, work 

#1
Jan910, 03:34 PM

P: 5

1. The problem statement, all variables and given/known data
A 11.0 kg crate is pulled up a rough incline with an initial speed of 1.3 m/s. The pulling force is 121.0 N parallel to the incline, which makes an angle of 12.6◦ with the horizontal. The coefficient of kinetic friction is 0.31 and the crate is pulled a distance of 8.0 m. The acceleration of gravity is 9.81 m/s^2 . 2. Relevant equations a) Find the work done by Earth’s gravity on the crate. Answer in units of J. 3. The attempt at a solution Fnormal=m*g*costheta Fnormal=105.3111778N Fnormal* coefficient of friction=Ffriction Ffriction=32.64646511N FappliedFfriction=Fnet Fnet=88.35353489N Work=N*m Work=706.8282791N*m (Work*sin12.6)=W done by gravity Work done by gravity is 154.1898119N*m I think I did it right, but the computer marked me wrong. Can anyone explained this? 



#2
Jan910, 04:00 PM

P: 621





#3
Jan910, 04:24 PM

P: 5

The displacement is parallel to the plane. If thats what you are asking




#4
Jan910, 05:17 PM

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Work done by gravity on a object on an inclined plane
Your last step is incorrect. I'm not sure what you're trying to do there. Can you explain your thinking?




#5
Jan910, 05:19 PM

P: 5

a) Find the work done by Earth’s gravity on the crate. Answer in units of J.
I think you have to find the work done on the block parallel to the plane, then find the y component of the work 



#6
Jan910, 05:42 PM

P: 621

Work is Force x displacement said very simply.
So the displacement is 8 m up and parallel the plane. All you need is the force due to gravity that is also parallel to the plane, thus also parallel to the displacement. What is the force due to gravity that is parallel to the displacement?And then, is it in the same direction as the displacement (+) work, or is it in the opposite direction as the displacement, () work. 



#7
Jan910, 09:18 PM

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When I said earlier that your steps were correct, I didn't mean to imply that they would all be necessarily useful in trying to calculate the work done by gravity, just that you didn't make any physics mistakes in deriving them. 



#8
Jan1010, 04:11 PM

P: 5

a) Find the work done by Earth’s gravity on the crate.
Weight*Distance sine theta=Work done by the gravity b) Find the work done by the force of friction on the crate. Weight*Distance cos theta*coefficient=Work done by the force of gravity c) Find the work done by the puller on the crate. Applied Force *Distance=Work done by the puller on the crate d) Find the change in kinetic energy of the crate. Answer to c(Answer to b+Answer to c)=The change in kinetic energy of the crate e) Find the speed of the crate after it is pulled 8.0 m. The square root of [((2*Answer to d/Mass)+Vi^2)] I finally figured c out and ask my classmate for the others http://answers.yahoo.com/question/in...2140953AATtl7k 



#9
Jan1010, 04:21 PM

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You should have negative answers for (a) and (b).




#10
Jun310, 04:08 PM

P: 5

For part d, it's Answer to c(Answer to b+Answer to a)=The change in kinetic energy of the crate




#11
Feb2811, 08:38 PM

P: 317

I believe that the amount of work done by gravity is equal to the potential gravitational energy. Therefor, if you find the height as it goes up a 8 meter incline, then you can also find the work using mgh (meters times gravitational acceleration times height). so I think the answer is 188 J, since sin12.5=height/8 where height is 1.745 meters, g=9.81m/s^2, and 11kg=mass. multiply them and you get the amount of joules it applied which should be 188 J. Well, then again if we consider the chnage in potential energy we can safely assume that the initial potential energy is 0. so, 188 j still holds true.



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