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?
The Attempt at a Solution
for part a, the sum of the forces in the x direction are 2ma=-Fk
where Fk=[tex]\mu[/tex]2mg the 2m's cancel and we are left with a=-[tex]\mu[/tex]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 im kind of having doubts for some reason
thanks in advance