If you use a horizontal force of 30.0 N to slide a 12.0 kg wooden crate across a ﬂoor at a constant velocity, what is the coefficient of friction between the crate and the floor?

Question

If you use a horizontal force of 30.0 N to slide a 12.0 kg wooden crate across a ﬂoor at a constant velocity, what is the coefficient of friction between the crate and the floor?

1 Answer

Impact of this question

73857 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-04T21:50:51

I found: $μ_{k} = 0.25$ $mu_k=0.25$

Explanation:

Have a look:
enter image source here
Newton's law is split into two components along $x$ $x$ and $y$ $y$ , that, using our data, give:
$30 - μ_{k} R_{n} = 0$ $30-mu_kR_n=0$
$R_{n} = m g = 12 \cdot 9.81 = 117.7 \approx 118 N$ $R_n=mg=12*9.81=117.7~~118N$
So:
$μ_{k} = \frac{30}{118} = 0.25$ $mu_k=30/118=0.25$

Science

Math

Humanities

... and beyond

If you use a horizontal force of 30.0 N to slide a 12.0 kg wooden crate across a ﬂoor at a constant velocity, what is the coefficient of friction between the crate and the floor?

1 Answer

Explanation:

Related questions

Impact of this question