# How much work would it take to push a  1 kg  weight up a  6 m  plane that is at an incline of  pi / 3 ?

Aug 20, 2017

$W \approx 51 \text{J}$

#### Explanation:

Work done by a constant force is given as the dot product of the force and displacement vectors:

$\textcolor{b l u e}{W = F \mathrm{dc} o s \theta}$

• Where $\theta$ is the angle between the force and displacement vectors. This is not the angle of incline.

Diagram: • Where ${\vec{F}}_{A}$ is the applied (pushing) force, $\vec{n}$ is the normal force, and ${\vec{F}}_{g}$ is the force of gravity, broken up into its parallel (x, horizontal) and perpendicular (y, vertical) components.

• I will define up the incline as positive

We can use Newton's second law to determine the net force on the object.

$\textcolor{b l u e}{{F}_{x \text{ net}} = \sum {F}_{x} = {F}_{A} - {F}_{G x} = m {a}_{x}}$

$\textcolor{b l u e}{{F}_{y \text{ net}} = \sum {F}_{x} = n - {F}_{G y} = m {a}_{y}}$

• We will assume dynamic equilibrium, as it is assumed that we are looking for the minimum amount of work required to push the object up the incline.

$\implies {F}_{A} - {F}_{G x} = 0$

$\implies n - {F}_{G y} = 0$

Which gives:

$\implies \textcolor{b l u e}{{F}_{A} = {F}_{G x}}$

$\implies \textcolor{b l u e}{n = {F}_{G y}}$

Therefore, the magnitude of the pushing force required is equal to the parallel component of the force of gravity. We can find this component using our diagram and basic trigonometry.

$\implies \sin \left(\theta\right) = \text{opposite"/"hypotenuse}$

$\implies \sin \left(\theta\right) = {F}_{G x} / {F}_{G}$

$\implies {F}_{G x} = {F}_{G} \sin \left(\theta\right)$

And as we know that ${F}_{G} = m g$:

$\implies \textcolor{b l u e}{{F}_{G x} = m g \sin \left(\theta\right)}$

Finally, we have:

${F}_{A} = m g \sin \left(\theta\right)$

Putting this into the equation for work:

$W = m g \sin \left({\theta}_{i}\right) \mathrm{dc} o s \left(\theta\right)$

Since the pushing force is in the same direction as the displacement (up the ramp), $\theta = 0$, and so we have:

$W = m g \sin \left(\theta\right) d$

We have the following information:

• $\mapsto m = 1 \text{kg}$
• $\mapsto d = 6 \text{m}$
• $\mapsto {\theta}_{i} = \frac{\pi}{3}$
• $\mapsto g = 9.81 {\text{m"//"s}}^{2}$

$\implies W = \left(1 \text{kg")(9.81"m"//"s"^2)(6"m}\right) \sin \left(\frac{\pi}{3}\right)$

$\implies \textcolor{b l u e}{W \approx 51 \text{J}}$