Monkey and sled - conservation of energy problem

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Homework Help Overview

The problem involves a monkey and a sled moving up an inclined track, where the conservation of energy principle is applied to determine how far they travel before coming to a stop. The scenario includes parameters such as initial speed, mass, incline angle, and friction coefficient, all of which are relevant to the analysis of energy transformations and forces acting on the system.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the conservation of energy equation, questioning how to calculate work done against friction and the normal force. There are attempts to clarify the roles of kinetic and potential energy, as well as the forces acting on the sled and monkey system.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the forces involved and the energy equation. Some guidance has been offered regarding the calculation of the normal force and the components of weight acting on the incline, but no consensus has been reached on the final approach to solving for the distance traveled.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available and the methods they can use. There is a focus on understanding the relationships between forces and energy without providing direct solutions.

physicsstudent06
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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?
 
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First just a slight correction;
physicsstudent06 said:
KEi + PEi + Wnc = KEf +PEf
This should read; KEi + PEi = KEf + PEf + 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 = KEi - PEf. 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?
 
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?
 
Start by finding the friction force acting on the sled. (Identify all the forces acting on the "monkey/sled" system.)
 
but d is still unknown
 
F=u Fn (u being the coefficient of friction), F being the normal force
 
If you set up your energy equation properly, "d" will be the only unknown. To find d, solve that equation.
 
physicsstudent06 said:
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;

[tex]dR\mu = 160 - 196d\sin(20) \Leftrightarrow d(R\mu + 196\sin(20)) = 160[/tex]

Can you go from here?
 
normal force = mass x acceleration due to gravity?
 
  • #10
physicsstudent06 said:
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.
 
  • #11
what is R?
 
  • #12
physicsstudent06 said:
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.
 
  • #13
i don't know how to find the friction force. force = ma, but how can i find the acceleration?
 
  • #14
physicsstudent06 said:
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;

[tex]F = \mu R[/tex]

Where R is the normal reaction force
 
  • #15
how do you find the normal reaction force?
 
  • #16
physicsstudent06 said:
how do you find the normal reaction force?
As Doc Al suggested draw a free body diagram of all the forces acting.
 
  • #17
anyone there?
 
  • #18
oh sorry, comp glitch
 
  • #19
physicsstudent06 said:
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.
 
  • #20
i'm not sure what other forces are acting. force of gravity, you said no.

force of sled monkey?
 
  • #21
i'm not sure what you mean by resolve all forces
 
  • #22
xxxxxxxxxxxx
 
  • #23
you there?
 
  • #24
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
 
  • #25
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.
 
  • #26
i drew a diagram, forces = gravity, friction, and whatever pushed the sled?
 
  • #27
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.
 
  • #28
physicsstudent06 said:
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 [itex]f = \mu N[/itex], 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.
 
  • #29
cos theta?
 
  • #30
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
 

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