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

In summary, the conversation discusses solving three physics problems involving friction and accelerations. The first problem involves calculating the force applied to a sled given its mass and coefficient of kinetic friction. The second problem involves finding the coefficient of kinetic friction for a snowboarder sliding down a slope with a given acceleration. The third problem involves finding the coefficient of kinetic friction between a baseball player and the ground as he slides to a stop at third base. The conversation includes discussions on drawing free body diagrams and using equations such as F = ma to solve for the unknown variables.
  • #1
Shinster007
17
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=[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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  • #2


Have you tried drawing the free body diagram? The sled is moving at constant speed so the net force acting on it is zero.
 
  • #3


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?
 
  • #4


Yes.
 
  • #5


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?
 
  • #6


Oops, *centripetal
 
  • #7


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]
 
  • #8


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
 
  • #9


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.
 

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

What is friction and how does it affect objects?

Friction is the force that opposes the motion of an object when it comes into contact with another surface. It acts in the opposite direction of an object's motion and can either slow down or prevent an object from moving. The amount of friction depends on the type of surfaces in contact and the force pushing them together.

How is tension related to physics?

Tension is a force that is transmitted through a string, rope, or cable when it is pulled tight. In physics, tension is often used in the context of analyzing the forces acting on a system. It can be calculated using Newton's second law of motion, which states that the net force on an object is equal to its mass times its acceleration.

What is Hooke's Law and how is it used in physics?

Hooke's Law states that the force required to extend or compress a spring by a certain distance is directly proportional to that distance. This means that the more the spring is stretched or compressed, the greater the force applied to it. Hooke's Law is used in physics to calculate the restoring force of a spring and its resulting motion.

How does centrifugal force work?

Centrifugal force is a fictitious force that appears to act on an object moving in a circular path. It is the tendency of an object to move away from the center of rotation and is caused by the inertia of the object. As the object moves in a circular path, it wants to continue moving in a straight line, resulting in the perceived outward force.

How do these concepts apply to real-life situations?

Friction, tension, springs, and centrifugal force are all important concepts that play a role in our daily lives. For example, friction is what allows us to walk without slipping and is also harnessed in tools such as brakes and grip tape. Tension is used in structures like bridges and buildings to distribute weight and maintain stability. Springs are found in everyday objects like mattresses, car suspensions, and door hinges. Centrifugal force is used in amusement park rides and is also essential in the functioning of centrifuges used in medical and scientific research.

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