Could someone Please help me

  • Thread starter Resmo112
  • Start date
  • #1
45
0

Homework Statement


A 15.0 kg box is pushed up an 8.00 m long frictionless incline inclined at an angle of 30.0 degrees above the horizontal in 4.2 seconds. Find the average power required.


Homework Equations


W=F*Deltad*CosTheta
Power= W/T
F=m*a
KE=1/2mv^2




The Attempt at a Solution


This isn't so much a homework problem it's an old exam problem that I can't for the life of me get. The answer key from my instructor says the answer is 140 W. I Have a weight a distance and a time, so I figure I can take my mass and gravity and get the total force, and then I should be able to take the distance time Cos(30) and multiply that by 147N to get the work put that over the time and that should be the average power. but it comes out to 198 and that's not right. Any help is greatly appreciated.
 

Answers and Replies

  • #2
283
0
The work is equal to the change in gravitational potential energy which is:

[tex] W = \Delta G.P.E = mg \Delta h = mg * (8 meters) * sin(30) [/tex]

dividing this by the time gives the correct answer.

Does this make sense?

I'm not sure that I understand how you have tried to solve it. If you would like to show each step of what you have done, with equations, I would be glad to show you where you have gone wrong.
 
  • #3
45
0
yeah that gives me the right answer. which I think I tried to do during this exam, and I don't remember getting the right answer.

Ok what I did was I tried to find the force of the crate so that's 15Kg*9.8=147 I actually did that but instead of sin i used cos because of the equation on my equation sheet W=F*d*cos(theta) and I still don't understand why you would use sin instead of cos. WAIT! I might have something here. because you're not trying to find the distance traveled by the block you're trying to find the height the block finishes at! which would be the opposite and not the adjacent. am I right in thinking this?
 
  • #4
283
0
yeah that gives me the right answer. which I think I tried to do during this exam, and I don't remember getting the right answer.

Ok what I did was I tried to find the force of the crate so that's 15Kg*9.8=147 I actually did that but instead of sin i used cos because of the equation on my equation sheet W=F*d*cos(theta) and I still don't understand why you would use sin instead of cos. WAIT! I might have something here. because you're not trying to find the distance traveled by the block you're trying to find the height the block finishes at! which would be the opposite and not the adjacent. am I right in thinking this?
Yes you are correct. The change in height is

[tex] \Delta h = d sin \theta [/tex]

However what you are doing is not correct. Are you saying that the only force acting on the crate is gravity? No, something must be pushing the crate up the incline, and the work done by that force is what we are looking for. Since the force is not given, the only way to find the work done by it is to consider the following.

You know that the change in mechanical energy must be equal to the work done by an external force. Now, we are to assume that the kinetic energy of the crate does not change during the displacement. But, we see that the block has increased in height, telling us that the gravitational potential energy has increased. Now we know that the pushing force has caused this change in gravitational energy. So if we find the change in energy, we know it must be equal to the work done. So you use this to find the work done by the pushing force, as was shown above.
 
  • #5
242
0
yes, you have it now. it is a good idea to be cautious anytime you use a formula with an angle in it. make sure you draw the picture, because you may be thinking of a different angle than the author of the formula was thinking of.

in this case, the formula [itex]W=F d cos(\theta)[/itex] comes from a vector dot product, and the angle should be that between the vectors F and d. But F and d both point up the incline, so the angle between them is zero, so [itex]W=Fd[/itex].

but, the force we need to apply is not equal to the weight of the cart, it only has to counter the component of the weight pointed down the incline, which is [itex]m g sin(\theta)[/itex]. so we get the formula [itex]W=m g d sin(\theta)[/itex]

hope this helps
 
  • #6
45
0
ok but according to the equation you posted, work is change in gravitational potential energy. which goes from 0 because there was no height to 147 *sin30*8 so if you did final minus initial you'd get the same number. IF the question had said something like you have a block 3 feet above the ground you push it up an inclined surface at 30 degrees 8 meters long for 4.2 seconds then you'd want MGH_f - MGH_i per time am I correct?
 
  • #7
283
0
ok but according to the equation you posted, work is change in gravitational potential energy. which goes from 0 because there was no height to 147 *sin30*8 so if you did final minus initial you'd get the same number. IF the question had said something like you have a block 3 feet above the ground you push it up an inclined surface at 30 degrees 8 meters long for 4.2 seconds then you'd want MGH_f - MGH_i per time am I correct?
Yes you are. But actually you can always set G.P.E = 0 at the initial point in a problem like this. It is an arbitrary decision where you choose to put G.P.E. = 0. It does not always have to be zero at ground level or sea level, its up to you.
 
  • #8
45
0
Thank you AlexChandler and Eczeno you've both been great help I really appreciate it.
 
  • #9
242
0
cheers
 

Related Threads on Could someone Please help me

Replies
7
Views
894
  • Last Post
Replies
1
Views
1K
Replies
3
Views
488
Replies
4
Views
1K
Replies
6
Views
6K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
945
Replies
1
Views
4K
Top