# Question d7d47

Dec 13, 2017

The $25 k g$ sack lifted $4 m$

#### Explanation:

Given: $50 k g$ sack lifted vertically a distance of $2 m$, and a $25 k g$ sack lifted vertically a distance of $4 m$

Find the mechanical work done in each case.

$\text{Work} = F \cdot d$ provided the force acts in the opposite direction as the object's motion. Where $\text{Force} \left(F\right)$ is measured in Newtons ($N$) and distance $\left(d\right)$ is measured in meters (m#)

The units of $\text{Work}$ is the $\text{Joule} \left(J\right)$

$1 J = 1 N \cdot m = \frac{1 k g \cdot {m}^{2}}{s} ^ 2$; $\text{ } 1 N = \frac{1 k g \cdot m}{s} ^ 2$

The weight of each sack exerts a force due to gravity: $W = m \cdot g$, where $m = \text{mass} \left(k g\right)$ and $g = 9.8 \frac{m}{s} ^ 2$

${F}_{1} = 50 \cdot \left(9.8\right) = 475 \frac{k g \cdot m}{s} ^ 2 = 475 N$

${F}_{2} = 25 \cdot \left(9.8\right) = 245 \frac{k g \cdot m}{s} ^ 2 = 245 N$

${W}_{1} = 475 \cdot 2 = 950 N \cdot m = 950 J$

${W}_{2} = 245 \cdot 4 = 980 N \cdot m = 980 J$

The $25 k g$ sack lifted $4 m$ requires more energy.