# How much work would it take to horizontally accelerate an object with a mass of 8  kg to 5  ms^-1 on a surface with a kinetic friction coefficient of 5 ?

Feb 17, 2016

$W = m d \left(a + \mu g\right) = 8 \cdot \frac{25}{6} \cdot \left(3 + 5 \cdot 9.8\right) = 1733.3$ $J$

See the explanation below for how this expression was derived.

#### Explanation:

(just a quick side note that a friction coefficient of $5$ is not very realistic: in the real world, coefficients of friction are typically between $0$ and $1$, though in some extreme cases they may rise as high as $2$)

There are two components of the force required. The first overcomes the inertia of the object in accordance with Newton's Second Law :

${F}_{1} = m a$

The second overcomes the frictional force:

${F}_{2} = \mu {F}_{N}$ where ${F}_{N}$ is the normal force between the object and the surface, which in this case is the weight force of the object, $F = m g$.

So the total force is:

${F}_{1} + {F}_{2} = m a + \mu m g = m \left(a + \mu g\right)$

The work done is the force exerted times the distance travelled:

$W = F d = m d \left(a + \mu g\right)$

Of the things in this expression we are given $m = 8$ $k g$, $\mu = 5$ (no units) and we know $g = 9.8$ $N k {g}^{-} 1$ (also $m {s}^{-} 2$ but this unit, which is exactly equivalent, makes more sense in this context).

We're told a final velocity of $v = 5$ $m {s}^{-} 1$ and are not told but can assume that the object starts at rest, $u = 0$ $m {s}^{-} 1$. We don't have values for $a$ $m {s}^{-} 2$ or $d$ $m$.

If we assume a value for the acceleration, $a$, that will also give us a value for the distance taken, $d$, which will supply the two remaining pieces of information we need. Assume $a = 3$ $m {s}^{-} 2$.

${v}^{2} = {u}^{2} + 2 a d$

Remembering $u = 0$ and rearranging:

$d = {v}^{2} / \left(2 a\right) = \frac{25}{2 \cdot 3} = \frac{25}{6}$ $m$

We can substitute these values into our expression for the work:

$W = m d \left(a + \mu g\right) = 8 \cdot \frac{25}{6} \cdot \left(3 + 5 \cdot 9.8\right) = 1733.3$ $J$

Here's an extension question for you: if we assumed a different number for the acceleration, would the work done be the same? Why or why not? Answer (write it down so you can't change your mind until you find out!) and then try it with $a = 10$ or some other value. Can you explain what you find?