Block being pushed along the floor (dynamics problem)

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SUMMARY

The discussion focuses on a dynamics problem involving a block sliding on the floor after a pushing force is removed. In part (a), when the mass of the block is doubled while maintaining the initial velocity, the distance slid before stopping remains unchanged due to the cancellation of mass in the acceleration equation. In part (b), when the initial velocity is doubled, the distance slid before stopping is quadrupled, as derived from the kinematic equation Vf² - Vi² = 2ad, confirming that acceleration remains constant.

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Homework Statement


a block pushed along the floor with velocity V0 slides a distance d after the pushing force is removed.
a) if the mass of the block is doubled but the initial velocity is not changed what is the distance the block slides before stopping?
b) if the initial velocity of the block is doubled to 2V0 but the mass is not changed, what is the distance the block slides before stopping?



Homework Equations





The Attempt at a Solution


for part a, the sum of the forces in the x direction are 2ma=-Fk
where Fk=\mu2mg the 2m's cancel and we are left with a=-\mug which is what the orginal acceleration was therefore the distance it travels is the same

for part b, since the mass has not changed we can assume that the acceleration is the same. using Vf2-Vi2= 2ad (where Vf is zero) we see that the distance is quadrupled

can some one check and see if i got this right I am kind of having doubts for some reason
thanks in advance
 
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Hi mjolnir80! :smile:

(have a mu: µ :wink:)
mjolnir80 said:
for part a, the sum of the forces in the x direction are 2ma=-Fk
where Fk=\mu2mg the 2m's cancel and we are left with a=-\mug which is what the orginal acceleration was therefore the distance it travels is the same

for part b, since the mass has not changed we can assume that the acceleration is the same. using Vf2-Vi2= 2ad (where Vf is zero) we see that the distance is quadrupled

Yes, that's fine! :smile:

You could also do it using the work-energy theorem …

try it that way, and see which way you prefer. :wink:
 

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