# Chapter 5 - Force and Motion-I - Problems: 2c

$\vec{a}=(3.0m/s^2)\vec{i}$

#### Work Step by Step

To find the acceleration, remember that $$\vec{F_{net}}=m\vec{a}$$ This means that the acceleration can be expressed as $$\vec{a}=\frac{\vec{F_{net}}}{m}$$ To find the net force, sum up all forces and their components. This yields a net force of $$\vec{F_{net}}=[(3.0N)\vec{i}+[(4.0N)\vec{j}]+(3.0N)\vec{i}+(-4.0N)\vec{j}]$$ $$\vec{F_{net}}=(6.0N)\vec{i}$$ Therefore, by substituting known values of $F_{net}=(6.0N)\vec{i}$ and $m=2.0kg$ yields an acceleration of $$\vec{a}=\frac{(6.0N)\vec{i}}{2.0kg}=(3.0m/s^2)\vec{i}$$

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