# How much work would it take to horizontally accelerate an object with a mass of 8 kg to 3 m/s on a surface with a kinetic friction coefficient of 5 ?

Feb 24, 2018

Recall,

$W = F \mathrm{dc} o s \theta$

If theta = 0°, then $W = F d$ (which is the case if the force is parallel to the displacement).

Apply Newton's second law to understand the amount of friction you need to overcome in order to push the particle,

$\Sigma {F}_{y} = {F}_{N} - m g = 0$

$\implies {F}_{N} = m g \approx 78.4 \text{N}$

Moreover, consider the situation at two instants: (i) at rest, and (ii) having been accelerated to (3"m")/"s".

Hence,

$0 = \frac{1}{2} m {\nu}^{2} - {\mu}_{k} \cdot {F}_{N}$

$\implies {\mu}_{k} \cdot {F}_{N} = \frac{1}{2} m {\nu}^{2}$

$\implies {\mu}_{k} \approx 0.5$ is the right kinetic friction coefficient (meanie)!

Moreover,

${\mu}_{k} \cdot {F}_{N} = {F}_{p} \approx 36 \text{N}$

Hence,

$W = F d \approx 36 \text{N} \cdot d$

where $d$ is your displacement (which you need to calculate work).