How do Newton's Laws work in these situations?

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Homework Help Overview

The discussion revolves around applying Newton's Laws to various scenarios involving friction, forces, and acceleration. Participants are examining specific problems related to a box being pulled on a surface with friction, a sled being pulled at an angle, and the net force acting on an object with changing velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to calculate friction forces and acceleration based on given forces and coefficients of friction. Questions arise regarding the application of static versus kinetic friction and the interpretation of results. One participant expresses uncertainty about starting a problem involving net force and changing velocity.

Discussion Status

Some participants have provided clarifications on terms related to friction and have pointed out the importance of using static friction in the context of the box problem. There is an ongoing exploration of different interpretations of the problems, particularly regarding the calculations and assumptions made.

Contextual Notes

There is a mention of confusion regarding the definitions of coefficients of friction and the distinction between static and kinetic friction. Some participants express limitations in their understanding of these concepts, which may affect their ability to engage with the problems fully.

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1) A 5kg box is pulled by a force F horizontally (no angle). There is friction between the box and horizontal surface. Mu_k = .25 and Mu_s = .5. If F=20N find the magnitude of the friction force and the acceleration of the block.

I used the formula |f_k| = mu_k * |Normal force|

f_k = .25 * 9.8 * 5
f_k = 12.25 N

The correct answer is that the friction is 20 N, meaning that the block didn't move. Where did I go wrong?

2) A 5kg sled is pulled an an angle 5 degree above the horizontal. Friction between the sled and the snow is described by mu_k = .1 and mu_s = .15. What is acceleration of the sled?

The correct answer is 2.99 m/s^2

This is what I did.

I took the 20 N and multiply by cos (5 degree) to get 19.92.

|f_k| = mu_k * N
f_k = .1*5*9.8
f_k = 4.9

(19.92 - 4.9) / 5 = a
a = 3.004

Did I just round my answer differently or did I do something wrong?

3) A 2kg object is under the influences of several forces. Assume that the net force is constant. At t=0, the object's velocity is 3i + 2j and at t=4s its velocity is 11i - 14j . Find the magnitude and direction of net force acting on the object. Please explain this to me. I have no idea how to start.

Thanks in advance for your help.
 
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Can you please tell me what Mu_k, Mu_s and |f_k| all mean? I'm doing similar stuff to this at the moment but I don't recognise those things you are using.
 
Mu_k is the coefficient of kinetic friction
Mu_s is coefficient of static friction

|f_k| is kinetic friction
|f_s| is static friction
 
Oh okay gotcha. Yeah I actually probably can't help because I haven't yet delved into static and kinetic friction really. I have worked with simply friction before without determining differences between the two
 
Before the box moves, it is static friction that is important. You calculated the acceleration using kinetic friction. Go back and do it using static friction.
 
Thanks for your help. Since the static friction is greater than the force pushing the object, the object will not move. Since 20N is applied on the object, the friction is also 20N according to Newton's Third Law. Am I correct?
 

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