Question

A 20Kg box is pushed across the floor at a constant velocity with a force of 200N. What is the

Answer 1

Answer:

The kinetic coefficient of friction between the box and the floor is 1.020.

Explanation:

Let suppose that box is on horizontal ground. According to the Newton's Laws, an object has a net acceleration of zero when either is at rest or moves at constant velocity. Friction is a reaction to the external force that moves the box. Hence, the equation of equilibrium for the 20-Kg box is:

$\Sigma F = F-f = 0$ (Eq. 1)

Where:

$F$ - External force, measured in newtons.

$f$ - Friction force, measured in newtons.

If we know that $F = 200\,N$, then the magnitude of the kinetic friction force is:

$f = F$

$f = 200\,N$

In addition, friction force is represented by the following formula:

$f = \mu_{k}\cdot m\cdot g$ (Eq. 2)

Where:

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

Now we clear the kinetic coefficient of friction:

$\mu_{k} = \frac{f}{m\cdot g}$

If we know that $f = 200\,N$, $m = 20\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$, then the kinetic coefficient of friction is:

$\mu_{k} = \frac{200\,N}{(20\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\mu_{k} = 1.020$

The kinetic coefficient of friction between the box and the floor is 1.020.