Object Traveling Up an Inclined Plane

  • #1
[SOLVED] Object Traveling Up an Inclined Plane

Homework Statement



In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 2.0 m/s up a 19.0° inclined track. The combined mass of monkey and sled is 16 kg, and the coefficient of kinetic friction between sled and incline is 0.20. How far up the incline do the monkey and sled move?


Homework Equations



Sigma(F) = ma



The Attempt at a Solution



I drew a free body diagram and found the normal force using Sigma(Fy)=ma.
Normal Force - Weight y = 0
Normal Force = Weight y
Normal Force = (weight)(cos 19)
Normal Force = 148.257 N

I can now find the Force of Friction:
Force of friction = coefficient of friction * Normal Force
Force of Friction = (.20)(148.257 N)
Force of Friction = 29.65 N (In the negative x direction.)

Now I'm stuck. I'm not sure how to apply Newton's laws since there is no force in the positive x direction. Any help is appreciated. Thanks.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
Well you could use energy considerations to find how far it travels up the plane. (Conservation of energy)
 
  • #3
I was thinking about using (mgh + .5mv^2)initial = (mgh + .5mv^2)final, but I thought this is only true for conservative forces.
 
  • #4
I believe I've solved it using energy. Here's what I did (My main equation is in bold):

Work (nonconservative) = (KE(f) + PE(f)) - (KE(i) + PE(i))
Work = F*Delta x

Substituting the latter into the former and getting rid of things that will be zero, I get:
Force*Delta x = (mgh(f)) - (.5mv(i)^2)

Now, using simple trigonometric relationships, I see that Sin theta = h/x. This gets plugged into my main equation, so:
Force*Delta x = (mg(x*sin theta)(f)) - (.5mv(i)^2)

I also know that my only nonconservative force is my friction force, so:
-Force friction*Delta x = (mg(x*sin theta)(f)) - (.5mv(i)^2)

I now have one equation and one unknown, and can solve. Plug in everything:
-29.651 N*Delta x = (16 kg)(9.8 m/s^2)(x*Sin(19)) - (.5)(16 kg)(2 m/s)^2

Simplified and units removed for clarification:
-29.651*Delta x = 51.04x - 32
-80.691x = -32
x = -32/-80.691


Finally:
x = .397 m

Does this look right to you all?
 
  • #5
Can anybody comment on whether or not this looks correct?
 
  • #6
737
0
This does look correct and is a clever way of solving the problem. More commonly, because the combined force against motion up the inclined plane is always constant (friction force you found + force from gravity), you can apply one of the four kinematic equations for constant acceleration.

Your original question was:
Now I'm stuck. I'm not sure how to apply Newton's laws since there is no force in the positive x direction.


When doing these problems, you usually change coordinate axes from the ordinary vertical is y and horizontal is x, to one where the direction of motion (up the plane) is x, and the normal force, which is perpendicular to that motion, is y.

You have done this by using cosine to find how much of the total force from gravity contributes to the normal force and used that to find the force opposing motion up the plane from friction. You also need to use sine to find the force from gravity that is directly opposing movement up the plane.

Then, try to move on from there.
 
  • #7
I see. Didn't even think to use kinematics.
 

Related Threads on Object Traveling Up an Inclined Plane

Replies
4
Views
6K
  • Last Post
Replies
7
Views
2K
Replies
24
Views
2K
  • Last Post
Replies
1
Views
4K
Replies
3
Views
10K
Replies
16
Views
13K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
2
Views
1K
Top