# How much work does it take to raise a 3 kg  weight 1 m ?

Dec 24, 2016

#### Answer:

The work done by the lift is $\approx 24 J$.

#### Explanation:

Work is given by $W = \vec{F} \Delta r \cos \left(\theta\right)$, where $\vec{F}$ is the applied force, $\Delta r$ is the displacement, and $\theta$ is the angle between the force and displacement vectors. Because the weight is moving upward and this is the direction that the force of the lift is applied, $\theta = {0}^{o}$, $W$ is positive, and $W = \vec{F} \Delta y$

The force that must be exerted to lift the weight is its physical weight, given by ${\vec{F}}_{g} = m g$. We are given $3 k g$ as a mass quantity, and $g$ is the free-fall acceleration, $9.8 \frac{m}{s} ^ 2$.

$\vec{F} = \left(3 k g\right) \left(9.8 \frac{m}{s} ^ 2\right) = 29.4 N$

Therefore, a force of $29.4 N$ was exerted on the weight over a vertical distance of $1 m$, and

$W = 29.4 N \cdot 1 m$

$W = 29.4 J \approx 24 J$

Note that this is the work done by the lift, not the net work done on the weight. The net work done on the weight is $0 J$, as the work done by the lift is equal and opposite the work done by gravity ($- 24 J$).