A crate 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.

Answer: d/2.