Monkey and sled - conservation of energy problem

In summary, the problem involves a monkey and sled with an initial speed of 4.0 m/s up a 20 degree inclined track. The combined mass of the monkey and sled is 20 kg, and the coefficient of kinetic friction between the sled and incline is 0.20. Using the conservation of energy equation, the work done against friction (Wnc) can be found by subtracting the initial kinetic energy from the final potential energy. To find the final potential energy, the height of the incline must be calculated using trigonometry. The forces acting on the system are weight, friction, and the normal force. The normal force can be found by balancing the component of weight perpendicular to the incline. The normal
  • #36
physicsstudent06 said:
so is the normal force equal to the weight force?
No. Re-read my previous post. (Pay attention to the word "component".)

Try this: A 1kg book lays on a horizontal table. What's the normal force? In this case, the normal force does equal the weight.

But what if the table is tilted? What happens to the normal force then?
 
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  • #37
ok, so the normal force would be 1 kg for that book.

in our scenario, normal force would equal weight times sin theta?
 
  • #38
or weight plus sin theta
 
  • #39
i don't know, I'm frustrated
 
  • #40
is the normal force mass x gravity x sin (theta)?
 
  • #41
sin(theta) would be the height over the hypotenuse. the hypotenuse is d, but we don't know the height
 
  • #42
gotta go, but thanks for your help so far
 
  • #43
physicsstudent06 said:
is the normal force mass x gravity x sin (theta)?
Here's another hint. The parallel & perpendicular (to the incline) components of the weight are:
[tex]W_\parallel = -mg \sin \theta[/tex]
[tex]W_\perp = - mg \cos \theta[/tex]
 
  • #44
Firstly let my clarify my comment which seems to be causing som confusion;
Hootenanny said:
Careful, the normal force always act perpendicular to the surface. Gravity in this case is not acting perpendicular to the surface.
I said, that gravity is not acting perpendicular to the plane, I never said that it is not acting at all! As Doc Al said, there will be a component(part) of gravity which is acting parallel to the plain and one which is acting perpendicular to the plane, you must take into account these forces. As Doc Al and I have said, I think that it would be useful to draw a diagram. Now, I have found a diagram that illustrates the compoents pretty well (I haven't got time to draw one myself). Note that the vin the image is not a force, it is simply illustrating the direction of the velocity.
http://img149.imageshack.us/img149/6381/inclinedplane8xd.jpg [Broken]
Taken from the PIRA website

So let us now take stock. There are X components of forces acting;
(1)Friction - parallel to and down the inclined plane.
(2)Normal Reaction force - perpendicular to the plane.

And we have two components of gravity; parallel ([itex]W_\parallel[/itex]) and perpendicular ([itex]W_\perp[/itex]), the equations for which Doc Al has supplied above. Now the component of weight acting parallel to the plane is acting downwards, in the opposite direction of the velocity and the same direction as the frictional force. The component of weight acting perpendicular to the surface is acting in the opposite direction to the normal reaction force (N in the above diagram). Now, as I said in a previous post, as the block is only accelerating in the plane parallel to the incline, this implies that the sum forces perpendicular to the plane must be zero. This implies that [itex]N = W_\perp[/itex]. Do you follow? Can you go from here?
 
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<h2>1. How does the monkey's weight affect the conservation of energy problem?</h2><p>The monkey's weight plays a crucial role in the conservation of energy problem. The heavier the monkey, the more energy is required to lift it up the sled. This means that the sled will have less energy available to reach the top of the hill. Conversely, a lighter monkey will require less energy and the sled will have more energy available to reach the top of the hill.</p><h2>2. Does the slope of the hill affect the conservation of energy problem?</h2><p>Yes, the slope of the hill does affect the conservation of energy problem. As the slope increases, the potential energy of the sled also increases. This means that the sled will require more energy to reach the top of the hill. On the other hand, a less steep slope will require less energy for the sled to reach the top.</p><h2>3. How does friction impact the conservation of energy problem in this scenario?</h2><p>Friction plays a significant role in the conservation of energy problem. As the sled moves up the hill, friction between the sled and the surface of the hill will cause a loss of energy. This means that the sled will require more energy to reach the top of the hill. To minimize the impact of friction, it is important to have a smooth surface and reduce the weight of the sled.</p><h2>4. Can the monkey's position on the sled affect the conservation of energy problem?</h2><p>Yes, the monkey's position on the sled can affect the conservation of energy problem. If the monkey is positioned towards the front of the sled, it will require more energy to lift the sled up the hill. However, if the monkey is positioned towards the back of the sled, it will provide a counterbalance and require less energy to reach the top of the hill.</p><h2>5. Is the conservation of energy problem affected by the speed of the sled?</h2><p>Yes, the speed of the sled does have an impact on the conservation of energy problem. If the sled is moving at a faster speed, it will require more energy to reach the top of the hill. This is because the kinetic energy of the sled will need to be converted into potential energy to reach the top. On the other hand, a slower speed will require less energy to reach the top of the hill.</p>

1. How does the monkey's weight affect the conservation of energy problem?

The monkey's weight plays a crucial role in the conservation of energy problem. The heavier the monkey, the more energy is required to lift it up the sled. This means that the sled will have less energy available to reach the top of the hill. Conversely, a lighter monkey will require less energy and the sled will have more energy available to reach the top of the hill.

2. Does the slope of the hill affect the conservation of energy problem?

Yes, the slope of the hill does affect the conservation of energy problem. As the slope increases, the potential energy of the sled also increases. This means that the sled will require more energy to reach the top of the hill. On the other hand, a less steep slope will require less energy for the sled to reach the top.

3. How does friction impact the conservation of energy problem in this scenario?

Friction plays a significant role in the conservation of energy problem. As the sled moves up the hill, friction between the sled and the surface of the hill will cause a loss of energy. This means that the sled will require more energy to reach the top of the hill. To minimize the impact of friction, it is important to have a smooth surface and reduce the weight of the sled.

4. Can the monkey's position on the sled affect the conservation of energy problem?

Yes, the monkey's position on the sled can affect the conservation of energy problem. If the monkey is positioned towards the front of the sled, it will require more energy to lift the sled up the hill. However, if the monkey is positioned towards the back of the sled, it will provide a counterbalance and require less energy to reach the top of the hill.

5. Is the conservation of energy problem affected by the speed of the sled?

Yes, the speed of the sled does have an impact on the conservation of energy problem. If the sled is moving at a faster speed, it will require more energy to reach the top of the hill. This is because the kinetic energy of the sled will need to be converted into potential energy to reach the top. On the other hand, a slower speed will require less energy to reach the top of the hill.

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