Monkey and sled - conservation of energy problem

[physicsstudent06]

I'm working on the following problem:

a monkey is strapped to a sled and both are given an initial speed of 4.0 m/s upa 20 degrees inclined track. The combined mass of monkey and sled is 20 kg, and the coefficient of kinetic friction between the sled and the incline is 0.20. How far up the incline do the monkey and sled move?

Using the conservation of energy equation, I knowthat
KEi + PEi + Wnc = KEf +PEf

Since both PEi and KEf are 0, the equation is as follows:

KEi+Wnc=PEf

how do i find Wnc?

 

[Hootenanny]

First just a slight correction;

KEi + PEi + Wnc = KEf +PEf

This should read; KE_i + PE_i = KE_f + PE_f + Wnc. The intial kinetic energy will be converted into potential energy and work, thus work and final potential energy should be on the same side of the equality. As you correctly say, the inital potential energy and final kinetic energy drops out leaving you with; Wnc = KE_i - PE_f. Now, you know the intial kinetic energy, you must next calculate the change in potential energy. For this you need to calculate the height. HINT: Trigonomentry. Can you go from here?

 

[physicsstudent06]

i know that KEi is 1/2 x 20 kg x 4.0 m/s^2, and PEf is 20kg x 9.8 m/s x sin(20) x d

right?

 

[Doc Al]

Start by finding the friction force acting on the sled. (Identify all the forces acting on the "monkey/sled" system.)

 

[physicsstudent06]

but d is still unknown

 

[physicsstudent06]

F=u Fn (u being the coefficient of friction), F being the normal force

 

[Doc Al]

If you set up your energy equation properly, "d" will be the only unknown. To find d, solve that equation.

 

[Hootenanny]

i know that KEi is 1/2 x 20 kg x 4.0 m/s^2, and PEf is 20kg x 9.8 m/s x sin(20) x d

right?

That is correct, so now you have;

dR\mu = 160 - 196d\sin(20) \Leftrightarrow d(R\mu + 196\sin(20)) = 160

Can you go from here?

 

[physicsstudent06]

normal force = mass x acceleration due to gravity?

 

[Hootenanny]

normal force = mass x acceleration due to gravity?

Careful, the normal force always act perpendicular to the surface. Gravity in this case is not acting perpendicular to the surface.

 

[physicsstudent06]

what is R?

 

[Doc Al]

F=u Fn (u being the coefficient of friction), F being the normal force

Right. So what's the normal force (Fn)?

Once you've got the the friction force, use it to express the "work" done against friction.

 

[physicsstudent06]

i don't know how to find the friction force. force = ma, but how can i find the acceleration?

 

[Hootenanny]

i don't know how to find the friction force. force = ma, but how can i find the acceleration?

For kinetic friction, the frictional force is simply given by;

F = \mu R

Where R is the normal reaction force

 

[physicsstudent06]

how do you find the normal reaction force?

 

[Hootenanny]

how do you find the normal reaction force?

As Doc Al suggested draw a free body diagram of all the forces acting.

 

[physicsstudent06]

anyone there?

 

[physicsstudent06]

oh sorry, comp glitch

 

[Hootenanny]

anyone there?

Draw a free body diagram. HINT: The sled is only moving parallel to the incline. Resolve all forces so that they are either parallel or perpendicular to the inclined plane.

 

[physicsstudent06]

i'm not sure what other forces are acting. force of gravity, you said no.

force of sled monkey?

 

[physicsstudent06]

i'm not sure what you mean by resolve all forces

 

[physicsstudent06]

xxxxxxxxxxxx

 

[physicsstudent06]

you there?

 

[physicsstudent06]

hey, can you just let me know if you're coming back? if you're working on something else too, that's cool. i just need to know if i should stick around

 

[Doc Al]

List all the forces acting on the monkey/sled and indicate their direction. A diagram is better (draw one for yourself), but start by making a list. Hint: I see 3 forces acting on this system.

 

[physicsstudent06]

i drew a diagram, forces = gravity, friction, and whatever pushed the sled?

 

[physicsstudent06]

i can't figure out how to find the force. W=dRu (distance, reaction force, coefficient of kinetic friction).

distance is the unknown we're solving for. coefficient for kinetic friciton is given (0.20). and r is ?

force is usually ma, but we don't have an acceleration.

 

[Doc Al]

i drew a diagram, forces = gravity, friction, and whatever pushed the sled?

Gravity (weight): Yes, where does it point? What's an expression for the weight of the monkey/sled?

Friction: Yes, where does it point? You know that f = \mu N, where N is the normal force.

Normal force: That's the one you missed. Where does it point? That's what we need to find in order to know the friction force.

"whatever pushed the sled": No... We are only studying the motion after the sled has been pushed. That force is no longer acting.

 

[physicsstudent06]

cos theta?

 

[physicsstudent06]

ok, weight is mass x gravity. that's already taken into consideration in the equation right?Wnc = .5 x 20kg x 4.0 m/s^2 - (20 kg)(9.8 m/s) sin (20) d

Wnc = dRu

 

[physicsstudent06]

d = unknown

u = 0.20

 

[physicsstudent06]

i still don't know how to find the normal force, or the force due to friction

 

[Doc Al]

To find the normal force, realize that it must exactly balance the component of the weight that is perpendicular to the incline. (The weight and normal force are the only two forces on the monkey/sled that have components perpendicular to the incline.)

 

[physicsstudent06]

weight force = 196 N

 

[physicsstudent06]

so is the normal force equal to the weight force?

 

[Doc Al]

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?

 

[physicsstudent06]

ok, so the normal force would be 1 kg for that book.

in our scenario, normal force would equal weight times sin theta?

 

[physicsstudent06]

or weight plus sin theta

 

[physicsstudent06]

i don't know, I'm frustrated

 

[physicsstudent06]

is the normal force mass x gravity x sin (theta)?

 

[physicsstudent06]

sin(theta) would be the height over the hypotenuse. the hypotenuse is d, but we don't know the height

 

[physicsstudent06]

gotta go, but thanks for your help so far

 

[Doc Al]

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:
W_\parallel = -mg \sin \theta
W_\perp = - mg \cos \theta

 

[Hootenanny]

Firstly let my clarify my comment which seems to be causing som confusion;

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
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 (W_\parallel) and perpendicular (W_\perp), 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 N = W_\perp. Do you follow? Can you go from here?