# On a flat surface a 25kg crate is pushed at a constant velocity with a force of 50N. Determine the coefficient of friction. What would the acceleration on the crate be if the applied force was doubled?

## and how far would the crate move in the 5 seconds if it began with a velocity of 2m/s?

Nov 2, 2016

$\textsf{\left(a\right)}$

$\textsf{\mu = 0.2}$

$\textsf{\left(b\right)}$

$\textsf{a = 2.0 \textcolor{w h i t e}{x} {\text{m/s}}^{2}}$

$\textsf{\left(c\right)}$

$\textsf{s = 35 \textcolor{w h i t e}{x} \text{m}}$

#### Explanation:

$\textsf{\left(a\right)}$

Since the crate is moving at constant velocity the net force on the crate is zero.

The frictional force due to the surface is given by:

$\textsf{F = \mu R}$

Since the crate is moving we are referring to the coefficient of dynamic friction.

$\textsf{R}$ is the reaction and is equal to the weight of the block mg.

$\therefore$$\textsf{R = m g = 25 \times 9.8 = 245 \textcolor{w h i t e}{x} N}$

sf(muRstackrel(color(white)(xxxxxxxxxxxxxx))(color(blue)(larr)Xstackrel(color(white)(xxxxxxxxxxxxxx))(color(blue)(rarr)$\textsf{50 \text{N}}$

The frictional force is balanced by the effort so:

$\textsf{50 = \mu \times 245}$

$\therefore$$\textsf{\mu = \frac{50}{245} = 0.2}$

$\textsf{\left(b\right)}$

The effort is now doubled so the forces on the block are not balanced.

sf(muRstackrel(color(white)(xxxxxxxxxxxxxx))(color(blue)(larr)Xstackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(rarr)$\textsf{100 \text{N}}$

The net force is given by:

$\textsf{F = 100 - \mu R}$

$\therefore$$\textsf{F = 100 - \left(245 \times 0.2\right) = 51 \textcolor{w h i t e}{x} N}$

Using Newton's Law:

$\textsf{F = m a}$

$\therefore$$\textsf{a = \frac{F}{m} = \frac{51}{25} = 2.0 \textcolor{w h i t e}{x} {\text{m/s}}^{2}}$

$\textsf{\left(c\right)}$

I will assume you mean how far the crate moves after the force has been doubled.

Using the equation of motion:

$\textsf{s = u t + \frac{1}{2} a {t}^{2}}$

Putting in the numbers:

$\textsf{s = \left(2 \times 5\right) + \frac{1}{2} \times 2 \times {5}^{2} = 35 \textcolor{w h i t e}{x} m}$