What Force is Needed to Push a 40kg Sled on Ice with μk = 0.05?

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Homework Help Overview

The discussion revolves around calculating forces related to friction in various scenarios, including a sled on ice, a snowboarder on a slope, and a baseball player sliding to a stop. The subject area primarily involves concepts from mechanics and friction.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss drawing free body diagrams and setting up equations based on forces acting on objects. There are attempts to calculate normal forces and frictional forces, as well as questions about the correctness of their approaches. Some participants express uncertainty about how to proceed with calculations involving acceleration and friction.

Discussion Status

The conversation includes various attempts to solve problems, with some participants providing guidance on using free body diagrams and equations. There is an ongoing exploration of different scenarios, and while some participants feel they are gaining understanding, there is no explicit consensus on all approaches.

Contextual Notes

Participants mention feeling uncertain about their understanding of physics concepts, particularly in relation to their homework assignments. There are indications of imposed homework rules, such as working with symbols before substituting numbers.

Shinster007
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Homework Statement


The first problem I am having trouble with is as follows:

A child pushes a 40kg sled across the ice at a constant speed. If μk = 0.05 calculate the force applied to the sled.


Homework Equations


I know that f=[itex]\mu[/itex]FN



The Attempt at a Solution



I know that the free body diagram would consist of a point with a direction in the x coordinate, but I have no idea how to go about this problem. Math and physics is not my strong subject; this is my first ever physics class
 
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Have you tried drawing the free body diagram? The sled is moving at constant speed so the net force acting on it is zero.
 


Here is what I came up with:

Ʃy=FN-mg=0. Therefore, FN=mg=(40kg)*(9.8 m/s2)=392N.

Ʃx=F-f=0. Therefore, F=f, which equals [itex]\mu[/itex]s*FN, which equals (0.05)*(392N), giving me an answer of 19.6N. Did I do this correctly?
 


Yes.
 


Okay thanks for the help, I am feeling kinda dumb now that I know it is that simple.

Next question:

A 60kg snowboarder accelerates down a 32 degree slope at 3.0m/s2. Calculate μk.

Attempt:

Obviously since there is an acceleration, I know that the forces in the x direction acting on the snowboarder do not equal zero. When setting up a free body diagram I have FN acting upward, Fg acting "downward" (but not on the y axis), then Fx in the positive x direction, and fx in the negative direction.

I then get the following:

Ʃy= FN-mg*cos∅=0;
so FN=(60kg*9.8m/s2)*(cos 32)=498N.

Ʃx=mg*sin∅-[itex]\mu[/itex]kFN=ma.

Is this correct so far? Would it then just be a matter of moving everything around algebraically and solving for [itex]\mu[/itex]k?
 


Oops, *centripetal
 


Shinster007 said:
Okay thanks for the help, I am feeling kinda dumb now that I know it is that simple.

Next question:

A 60kg snowboarder accelerates down a 32 degree slope at 3.0m/s2. Calculate μk.

Attempt:

Obviously since there is an acceleration, I know that the forces in the x direction acting on the snowboarder do not equal zero. When setting up a free body diagram I have FN acting upward, Fg acting "downward" (but not on the y axis), then Fx in the positive x direction, and fx in the negative direction.

I then get the following:

Ʃy= FN-mg*cos∅=0;
so FN=(60kg*9.8m/s2)*(cos 32)=498N.

Ʃx=mg*sin∅-[itex]\mu[/itex]kFN=ma.

Is this correct so far? Would it then just be a matter of moving everything around algebraically and solving for [itex]\mu[/itex]k?

Yes, simply solve for [itex]\mu_k[/itex]
 


Okay, getting the hang of this.

Last question:

A baseball player initially running at 3.4m/s slides to a stop at third base in 1.2 seconds. Calculate the μk between him and the ground.

Attempt:

I'm not sure how to approach this problem. 3.4m/s divided by the 1.2s will give me the acceleration I think (or I guess deceleration in this case). Other than that I am at a loss
 


Shinster007 said:
Okay, getting the hang of this.

Last question:

A baseball player initially running at 3.4m/s slides to a stop at third base in 1.2 seconds. Calculate the μk between him and the ground.

Attempt:

I'm not sure how to approach this problem. 3.4m/s divided by the 1.2s will give me the acceleration I think (or I guess deceleration in this case). Other than that I am at a loss

The frictional force provides the negative acceleration, necessary for him to come to a stop. Using the negative acceleration calculated above, you can find [itex]\mu_k.[/itex]
 
  • #10


Well I know that f=[itex]\mu[/itex]k*FN, but I don't have a mass in order to calculate FN, nor do I have f. What equation can I use to solve the problem?
 
  • #11


Shinster007 said:
Well I know that f=[itex]\mu[/itex]k*FN, but I don't have a mass in order to calculate FN, nor do I have f. What equation can I use to solve the problem?
Simply use [itex]F = ma.[/itex] You know the only force acting horizontally on the baseball player as he slides (force of friction). This is your [itex]F.[/itex] Now sub in what [itex]F[/itex] is equal to and what do you notice about [itex]m[/itex]?
 
  • #12


Shinster007 said:
Well I know that f=[itex]\mu[/itex]k*FN, but I don't have a mass in order to calculate FN, nor do I have f. What equation can I use to solve the problem?
Your teacher probably encourages you to work problems out using symbols first and plug numbers in at the end. This is one of the reasons why.
 

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