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?

Question

Physics

Jaylon Williams

9 January, 04:42

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?

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Answers (1)

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Joseph Eaton · Answer 1

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.