Acrate pushed along the floor with velocity v⃗ i slides a distance d after the pushing force is removed. if the mass of the crate is doubled but the initial velocity is not changed, what distance does the crate slide before stopping?

Define
m = the  mass of the crate.
F = μmg, the resistive dynamic frictional force,
      where μ =dynamic coefficient of friction.
a =  the deceleration, when the crate slides subject only to to the frictional force.

The crate travels a distance d with initial velocity of v. Therefore
v² - 2ad = 0
a = v²/(2d)            (1)

Also,
F = ma
F  = (mv²)/(2d)     (2)

When m is doubled, then
F = (2mv²)/(2d) = (mv²)/d
The corresponding deceleration is
a = F/m = v²/d

Therefore, the new distance traveled, D, is given by
v² - 2(v²/d)D = 0
D = d/2
The new distance traveled is one half of d.

d/2.