An object with a mass of $2 k g$ $2 kg$ is acted on by two forces. The first is $F_{1} = < - 6 N, 2 N >$ $F_1= < -6 N , 2 N>$ and the second is $F_{2} = < 1 N, 3 N >$ $F_2 = < 1 N, 3 N>$ . What is the object's rate and direction of acceleration?

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Answer 1 · 2017-02-23T17:43:25

The net acceleration is $= 3.54 m s^{- 2}$ $=3.54ms^-2$ in the direction of $= (\frac{3}{4} π) r a d$ $=(3/4pi)rad$

The resultant force is

$\to F = {\to F}_{1} + {\to F}_{2}$ $vecF=vecF_1+vecF_2$

$= < - 6, 2 > + < 1, 3 >$ $= <-6,2>+<1,3>$

$= < - 5, 5 >$ $=<-5,5>$

$\to F = m \to a$ $vecF=m veca$

$\to a = \frac{1}{m} \cdot \to F$ $veca =1/m*vecF$

$= \frac{1}{2} < - 5, 5 > = < - \frac{5}{2}, \frac{5}{2} >$ $=1/2<-5,5> = <-5/2, 5/2>$

The magnitude of the acceleration is

$∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ < - \frac{5}{2}, \frac{5}{2} > ∣ ∣ ∣ ∣ ∣ ∣$ $||veca||=||<-5/2,5/2>||$

$= \sqrt{{(- \frac{5}{2})}^{2} + {(\frac{5}{2})}^{2}}$ $=sqrt((-5/2)^2+(5/2)^2)$

$= \sqrt{\frac{25}{4} + \frac{25}{4}}$ $=sqrt(25/4+25/4)$

$= \frac{\sqrt{50}}{2} = 3.54 m s^{- 2}$ $=sqrt 50/2=3.54ms^-2$

The direction is $θ = arctan (\frac{\frac{5}{2}}{- \frac{5}{2}})$ $theta = arctan((5/2)/(-5/2))$

The angle is in the 2nd quadrant

$θ = arctan (- 1) = (\frac{3}{4} π) r a d$ $theta=arctan(-1)=(3/4pi) rad$

An object with a mass of 2kg2 kg is acted on by two forces. The first is F1=<−6N,2N>F_1= < -6 N , 2 N> and the second is F2=<1N,3N>F_2 = < 1 N, 3 N>. What is the object's rate and direction of acceleration?