Question

A block pushed along the floor with velocity $v_{\rm 0x}$ slides a distance after the pushing force is removed.

Part A

If the mass of the block is doubled but itsinitial velocity is not changed, what distance does the block slidebefore stopping?

Express your answer in terms of thevariable .

Part B

If the initial velocity is doubled to $2v_{\rm 0x}$ but the mass is not changed, what distance does theblock slide before stopping?

Express your answer in terms of thevariable .

Answer 1

1)

the friction force F = -mgμ = ma

=>a = -gμ

final speed v_f = 0,

v_f² -v₀² =0-v₀² =2ad = -2gμd

=> d = v₀² / (2gμ)

so d is independ of mass,

so, If the mass of the block is doubled but theinitial velocity is not changed

the distance the block slides before stopping is the samed.

2)

now v₀ ->2v₀

=>d1 = (2v₀)² / (2gμ) = 4d

the distance is four times of d.

Answer 2

d, since friction is proportional to mass m will not effect d

4d, 2x the speed is 4x the kinetic energy, so 4x the distance for friction to stop

Answer 3

a)The distance the block travels depends on the size of the acceleration that is slowing it

down. Doubling the mass doubles the inertia, but it also doubles the friction and so the

acceleration is unchanged. Thus the stopping distance is unchanged in this model.

b)An object travelling twice the speed but the same acceleration will take twice as long to

slow down. Since its average velocity is also twice as much it will travel four times the

distance before stopping.