Question #16241

A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.20 and the push imparts an initial speed of 4.0 m/s. I know the answer I just am lost as to how to solve it

Expert's answer

We have that

a = F/m,

with F just the force of friction Ff. But

Ff = Fn*u,

where Fn is the normal force:

Fn = m*g.

Therefore,

a = Fn*u/m = m*g*u/m,

and you don't need to know m, as

a = u*g.

We have

0 = 16[m²/s²] + 2*u*g*d

==> d = -16[m²/s²]/(2*(-9.8)[m/s²] * 0.2)

==> d ~ 4.1 [m].

a = F/m,

with F just the force of friction Ff. But

Ff = Fn*u,

where Fn is the normal force:

Fn = m*g.

Therefore,

a = Fn*u/m = m*g*u/m,

and you don't need to know m, as

a = u*g.

We have

0 = 16[m²/s²] + 2*u*g*d

==> d = -16[m²/s²]/(2*(-9.8)[m/s²] * 0.2)

==> d ~ 4.1 [m].

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