Block being pushed along the floor (dynamics problem)

[mjolnir80]

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=-F_k
where F_k=$$\mu$$2mg the 2m's cancel and we are left with a=-$$\mu$$g 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

 

[tiny-tim]

Hi mjolnir80!

(have a mu: µ )

for part a, the sum of the forces in the x direction are 2ma=-F_k
where F_k=$$\mu$$2mg the 2m's cancel and we are left with a=-$$\mu$$g 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!

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

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

 

[thomate1]

Your approach to solving the problem is correct. In part a, you correctly identified that the acceleration remains the same since the mass is doubled but the initial velocity is not changed. Therefore, the distance the block slides before stopping will also remain the same.

In part b, you correctly used the equation Vf^2 - Vi^2 = 2ad to find the distance the block slides before stopping. Doubling the initial velocity will result in a quadrupling of the distance traveled, as you have calculated.

Overall, your solution is correct and your understanding of the problem is clear. Great job!