# How much work would it take to horizontally accelerate an object with a mass of 4 kg to 2 m/s on a surface with a kinetic friction coefficient of 3 ?

Jul 19, 2017

$W = 8$ $\text{J}$

#### Explanation:

We're asked to find the work necessary to accelerate a $4$-$\text{kg}$ object horizontally from rest to $2$ $\text{m/s}$ on a surface where the coefficient of kinetic friction is $3$.

To do this, we'll use the equation

${f}_{k} = {\mu}_{k} n$

where

• ${f}_{k}$ is the magnitude of the retarding friction force

• ${\mu}_{k}$ is the coefficient of kinetic friction ($3$)

• $n$ is the magnitude of the normal force, which since the plane is horizontal is equal to

$n = m g = \left(4 \textcolor{w h i t e}{l} {\text{kg")(9.81color(white)(l)"m/s}}^{2}\right) = 39.24$ $\text{N}$

The kinetic friction force is thus

${f}_{k} = \left(3\right) \left(39.24 \textcolor{w h i t e}{l} \text{N}\right) = 118$ $\text{N}$

This means a force of at least $118$ $\text{N}$ is needed to keep the object accelerating. The force doesn't matter, because the larger the force, the shorter the displacement once the object reaches $2$ $\text{m/s}$.

The work will be the same regardless of the force here; let's say the applied force is $119$ $\text{N}$, so the net force is

$119$ $\text{N}$ $- 118$ $\text{N}$ $= 1$ $\text{N}$

The magnitude of the acceleration is

${a}_{x} = \frac{\sum {F}_{x}}{m} = \left(1 \textcolor{w h i t e}{l} \text{N")/(4color(white)(l)"kg}\right) = 0.25$ ${\text{m/s}}^{2}$

Using kinematics, we can find the displacement $\Delta x$ after it reaches $2$ $\text{m/s}$, using the equation

${\left({v}_{x}\right)}^{2} = {\left({v}_{0 x}\right)}^{2} + 2 {a}_{x} \left(\Delta x\right)$

It starts from rest, so we have

$\left(2 \textcolor{w h i t e}{l} {\text{m/s")^2 = 0 + 2(0.25color(white)(l)"m/s}}^{2}\right) \left(\Delta x\right)$

$\Delta x = 8$ $\text{m}$

Using

$W = F s$

we have

W = (1color(white)(l)"N")(8color(white)(l)"m") = color(red)(8 color(red)("N"·"m" = color(red)(8 color(red)("J"