Relationship between Newton's laws and distance

AI Thread Summary
The discussion focuses on the relationship between Newton's laws and the distance a crate slides before stopping under varying conditions of mass and velocity. When the mass of the crate is doubled while keeping the initial velocity constant, the distance it slides is calculated using the equation d = (initial velocity squared) (mass)/kinetic friction. Conversely, when the initial velocity is doubled while keeping the mass the same, the distance becomes d = (2 times initial velocity squared) (mass)/kinetic friction. Participants emphasize the importance of correctly applying the kinetic friction formula, which is related to the normal force and the mass of the crate. The conversation aims to clarify the calculations and ensure understanding of the underlying physics principles.
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Homework Statement



a)If the mass of the crate is doubled but the initial velocity is not changed, what distance does the crate slide before stopping? Explain.

b) If the initial velocity of the crate is doubled to 2vi but the mass is not changd, what distance does the crate slide before stopping? Explain.

Homework Equations



V2=Vi2 + 2a(Xf-Xi)


The Attempt at a Solution


in this case, since the motion is horizontal, I though the X COMPONENT OF THE normal force = 0 and the X COMPONENT of the weight is equal to zero. I tried the problem again and for the condition (2m,v) my new answer is d= (initial velocity squared) (mass)/kinetic friction. ANd for condition (m, 2v) my answer is d= (2 times initial velocity squared) (mass)/kinetic friction. I hope these are correct. Thank you kindly for your help. Also, I know this was posted before but I posted it as a reply and i just want to make sure that this is answered so that is why i started another threat. Sorry for taking up the space but can you please tell me if this is right?
 
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someone seriously please help!
 
sona1177 said:

Homework Equations



V2=Vi2 + 2a(Xf-Xi)
Good, that is the right equation to use to think about this.

The Attempt at a Solution


in this case, since the motion is horizontal, I though the X COMPONENT OF THE normal force = 0 and the X COMPONENT of the weight is equal to zero. I tried the problem again and for the condition (2m,v) my new answer is d= (initial velocity squared) (mass)/kinetic friction. ANd for condition (m, 2v) my answer is d= (2 times initial velocity squared) (mass)/kinetic friction. I hope these are correct. Thank you kindly for your help.
Yes, so far so good. Uh, except the "2" should be "1/2", but that won't really matter here.

Next step would be to work on the kinetic friction. What is that equal to? (Hint: use the relation between kinetic friction and the normal force.) Can you relate it to the mass of the crate?
 
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