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

Click For Summary
SUMMARY

The discussion focuses on calculating the force required to push a 40kg sled on ice with a coefficient of kinetic friction (μk) of 0.05. The applied force (F) is determined using the equation F = μk * FN, where FN is the normal force calculated as FN = mg, resulting in FN = 392N and F = 19.6N. Additionally, the conversation explores similar problems involving a snowboarder on a slope and a baseball player sliding to a stop, emphasizing the importance of free body diagrams and the relationship between forces and acceleration.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of free body diagrams
  • Familiarity with the concept of kinetic friction
  • Ability to perform basic algebraic manipulations
NEXT STEPS
  • Study the principles of Newton's second law (F = ma)
  • Learn how to draw and interpret free body diagrams
  • Explore the calculation of friction coefficients in various scenarios
  • Investigate the effects of angles on forces in inclined planes
USEFUL FOR

Students in introductory physics courses, educators teaching mechanics, and anyone interested in understanding the dynamics of motion involving friction and forces.

Shinster007
Messages
17
Reaction score
0

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=\muFN



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
 
Physics news on Phys.org


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 \mus*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∅-\mukFN=ma.

Is this correct so far? Would it then just be a matter of moving everything around algebraically and solving for \muk?
 


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∅-\mukFN=ma.

Is this correct so far? Would it then just be a matter of moving everything around algebraically and solving for \muk?

Yes, simply solve for \mu_k
 


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 \mu_k.
 
  • #10


Well I know that f=\muk*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=\muk*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 F = ma. You know the only force acting horizontally on the baseball player as he slides (force of friction). This is your F. Now sub in what F is equal to and what do you notice about m?
 
  • #12


Shinster007 said:
Well I know that f=\muk*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.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
4
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
7K