Change in Potential Energy

  • Thread starter Medgirl314
  • Start date
  • #1
561
2

Homework Statement


A 20 kg child slides down a slide at a playground. The slide is 5 meters long, and is inclined at an angle of 37 degrees above the horizontal. What is the child's change in potential?

Homework Equations


PE=mgy

With m=mass, g=gravity, and y=distance.


The Attempt at a Solution



I have solved potential energy problems before this,but I don't know how to account for the angle. If there was no angle, the formula would be PE=20*9.8*5, and I think that would be the change in energy since when the child reaches the bottom y=0, canceling out the entire equation. Could someone please tell me where the angle comes in? I don't want it done for me, I just need one post with a helpful explanation, and then I should be good. :)

Thanks so much!
 

Answers and Replies

  • #2
21,489
4,867

Homework Statement


A 20 kg child slides down a slide at a playground. The slide is 5 meters long, and is inclined at an angle of 37 degrees above the horizontal. What is the child's change in potential?

Homework Equations


PE=mgy

With m=mass, g=gravity, and y=distance.


The Attempt at a Solution



I have solved potential energy problems before this,but I don't know how to account for the angle. If there was no angle, the formula would be PE=20*9.8*5, and I think that would be the change in energy since when the child reaches the bottom y=0, canceling out the entire equation. Could someone please tell me where the angle comes in? I don't want it done for me, I just need one post with a helpful explanation, and then I should be good. :)

Thanks so much!
There are two ways of doing this that give the same answer.

Method 1: You use the equation you presented, but instead of the 5, you use the vertical distance that the child descended.

Method 2. You use the equation you presented, but instead of the 9.8, you use the tangential component of g along the slide.

Chet
 
  • #3
561
2
Thank you! I knew trig would come in somewhere, I just couldn't figure out how to bring it in. So using Method 1, would the vertical distance be 3.94 meters?
 
  • #4
21,489
4,867
Thank you! I knew trig would come in somewhere, I just couldn't figure out how to bring it in. So using Method 1, would the vertical distance be 3.94 meters?
No. Try again. Draw a diagram this time.

Chet
 
  • #5
561
2
I drew one, but I did the trig wrong, I think. Would 5tan38 work?

Thanks again!
 
  • #7
adjacent
Gold Member
1,553
63
attachment.php?attachmentid=67097&stc=1&d=1393579060.png


Find h.
You won't need me here(Adjacent side) :cry:
 

Attachments

  • trig.png
    trig.png
    875 bytes · Views: 403
  • #8
561
2
Hm, I'm not sure why I can't wrap my mind around this. It would have to be sine, but why? Wouldn't that be the two unknown sides?

Thanks!
 
  • #9
Doc Al
Mentor
45,204
1,543
Hm, I'm not sure why I can't wrap my mind around this. It would have to be sine, but why? Wouldn't that be the two unknown sides?
What's the definition of sine for a right triangle?
 
  • #11
561
2
Sine=opposite/hypotenuse

Oh! Obviously, 5 is the hypotenuse. I let adjacent's letter choice confuse me. So 5sin 37=3.0 m. Correct?
 
  • #13
Doc Al
Mentor
45,204
1,543
Sine=opposite/hypotenuse

Oh! Obviously, 5 is the hypotenuse. I let adjacent's letter choice confuse me. So 5sin 37=3.0 m. Correct?
Correct.
 
  • #14
561
2
PE=mgy
PE=20*9.8*3.0=588 Joules

Correct?

Thanks again!
 
  • #15
Doc Al
Mentor
45,204
1,543
PE=mgy
PE=20*9.8*3.0=588 Joules

Correct?
Good. That's the magnitude of the change, but what's the sign of the change?
 
  • #16
561
2
I forgot about this thread. Oops. The sign, as in negative or positive? It depends on what direction we decided is the positive direction, right? So if we decided that down was the positive direction, then h would be negative, so the answer would be negative? But can you have negative energy? Would I just say 588 Joules down?

Thanks!
 
  • #17
Doc Al
Mentor
45,204
1,543
The sign, as in negative or positive?
Right.

It depends on what direction we decided is the positive direction, right? So if we decided that down was the positive direction, then h would be negative, so the answer would be negative? But can you have negative energy? Would I just say 588 Joules down?
It's got nothing to do with direction (energy is not a vector). It has to do with whether the potential energy increases (a positive change) or decreases (a negative change).

Since the child slides down the incline, is her change in PE positive or negative?
 
  • #18
561
2
Her change in PE would be negative, so the KE would be positive.
 
  • #19
Doc Al
Mentor
45,204
1,543
Her change in PE would be negative, so the KE would be positive.
Right. Since the change in PE is negative (the PE decreases), the change in KE will be positive (the KE increases). (The KE itself is always positive--unless it's zero.)
 
  • #20
561
2
Thank you!
 

Related Threads on Change in Potential Energy

Replies
5
Views
5K
E
  • Last Post
Replies
3
Views
10K
Replies
9
Views
15K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
3
Views
594
  • Last Post
Replies
7
Views
4K
Replies
3
Views
5K
  • Last Post
Replies
1
Views
7K
Replies
2
Views
7K
Top