# How much work does it take to raise a 6 kg  weight 28 m ?

Aug 10, 2017

$W = 168 \text{J}$

#### Explanation:

The work done by a constant force is given by the dot product:

$W = \vec{F} \cdot \vec{d}$

Giving the familiar equation:

$\implies \textcolor{\mathrm{da} r k b l u e}{W = F \mathrm{dc} o s \left(\theta\right)}$

where $F$ is the force applied to the object, $d$ is the displacement of the object, and $\theta$ is the angle between the force and displacement vectors

In raising the object, there is only the force which is being applied to cause the lift and the force of gravity acting against it.

Therefore, we can write:

$\sum F = {F}_{a} - {F}_{G} = m a$

We'll assume dynamic equilibrium (no acceleration but motion), giving:

$F a = {F}_{G}$

$\therefore$ The applied force must be equal to the force of gravity (opposite in direction).

The force of gravity acting on a stationary object at or near the surface of the earth is given by:

${F}_{G} = m g$

Which gives a new equation for work:

$\textcolor{\in \mathrm{di} g o}{{W}_{\text{lift}} = m g \cos \left(\theta\right)}$

We are given the following information:

• $\mapsto m = 6 \text{kg}$

• $\mapsto d = 28 \text{m}$ (vertically)

• $\mapsto \theta = {0}^{o}$ (implied)

Therefore:

$W = \left(6 \text{kg")(28"m}\right)$

$= 168 \text{J}$

$- - - -$

Note that $\theta = 0$ because the applied force and displacement occur in the same direction: upward. Therefore the angle between the vectors is ${0}^{o}$.