# Question 84f65

Jan 11, 2018

$F \text{net} = \frac{20}{t} \left[N\right]$ or
$F \text{net} = \frac{600}{d} \left[N\right]$

Where [N] is the unit Newton, t = time duration and d= distance traveled by the force.

#### Explanation:

You can find the unbalanced force provided you have the additional information required, i.e., either the time or distance involved in the change :

${F}_{\text{net" =ma rArr F"net}} = m \frac{{v}_{2} - {v}_{1}}{t}$

${F}_{\text{net}} = \frac{2 k g \left(20 \frac{m}{s} - 10 \frac{m}{s}\right)}{t} = \frac{20}{t} \left(k g \frac{m}{s}\right)$

or you can use work energy principle,
${W}_{\text{net"= DeltaKrArr F_"net}} d = \Delta K$
where
${W}_{\ne} t =$ net work
${F}_{\text{net}} =$ unbalanced net force,
$d =$ distance traveled by the force
$K =$ kinetic energ$y = \frac{1}{2} m {v}^{2}$
$\Delta K = {K}_{2} - {K}_{1}$

Hence,
${F}_{\text{net" =( DeltaK)/F_"net}} = \frac{\frac{1}{2} {m}_{2} {v}^{2} - \frac{1}{2} m {v}_{1}^{2}}{d}$
F_"net" =½(4kg)((20m)^2-(10m)^2)/d = 600J/d#