An object with a mass of $4$ $4$ $k g$ $kg$ is acted on by two forces. The first is $F_{1} = < - 4$ $F_1=<-4$ $N, 5$ $N, 5$ $N >$ $N>$ and the second is $F_{2} = < 2$ $F_2 = <2$ $N, - 8$ $N, -8$ $N >$ $N>$ . What is the object's rate and direction of acceleration?

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An object with a mass of $4$ $4$ $k g$ $kg$ is acted on by two forces. The first is $F_{1} = < - 4$ $F_1=<-4$ $N, 5$ $N, 5$ $N >$ $N>$ and the second is $F_{2} = < 2$ $F_2 = <2$ $N, - 8$ $N, -8$ $N >$ $N>$ . What is the object's rate and direction of acceleration?

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Answer 1 · 2016-08-28T04:27:13

The object is accelerating at $0.90 {m/s}^{2}$ $0.90 "m/s"^2$ in the direction ${56.3}^{o}$ $56.3^"o"$ counterclockwise from negative $x$ $x$ -axis.

Explanation:

The net force, $- - \to F_{net}$ $vec(F_"net")$ , is given by

$- - \to F_{net} = - \to F_{1} + - \to F_{2}$ $vec(F_"net") = vec(F_1) + vec(F_2)$

$= < - 4 N, 5 N > + < 2 N, - 8 N >$ $= < -4 "N", 5 "N" > + < 2 "N", -8 "N" >$

$= < - 2 N, - 3 N >$ $= < -2 "N", -3 "N" >$

Recall Newton's $2^{nd}$ $2^"nd"$ law

$- - \to F_{net} = m \to a$ $vec(F_"net") = m vec(a)$

where $m$ $m$ is the mass of an object and $\to a$ $vec(a)$ is its acceleration.

Substituting the values in, the resulting equation is

$< - 2 N, - 3 N > = (4 kg) \cdot < a_{x}, a_{y} >$ $< -2 "N", -3 "N" > = (4 "kg") * < a_"x" , a_"y" >$

Solving for $a_{x}$ $a_"x"$ and $a_{y}$ $a_"y"$ ,

$a_{x} = \frac{- 2 N}{4 kg} = - 0.5 {m/s}^{2}$ $a_"x" = frac{-2 "N"}{4 "kg"} = -0.5 "m/s"^2$
$a_{y} = \frac{- 3 N}{4 kg} = - 0.75 {m/s}^{2}$ $a_"y" = frac{-3 "N"}{4 "kg"} = -0.75 "m/s"^2$

The object's acceleration is found to be

$\to a = < - 0.5 {m/s}^{2}, - 0.75 {m/s}^{2} >$ $vec(a) = < -0.5 "m/s"^2, -0.75 "m/s"^2 >$

Its magnitude is

$∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣ = \sqrt{{(- 0.5 {m/s}^{2})}^{2} + {(- 0.75 {m/s}^{2})}^{2}}$ $||vec(a)|| = sqrt((-0.5 "m/s"^2)^2 + (-0.75 "m/s"^2)^2)$

$= 0.90 {m/s}^{2}$ $= 0.90 "m/s"^2$

Its direction is

${tan}^{- 1} (∣ ∣ ∣ ∣ \frac{- 0.75 {m/s}^{2}}{- 0.5 {m/s}^{2}} ∣ ∣ ∣ ∣) = {56.3}^{o}$ $tan^{-1}(abs(frac{-0.75 "m/s"^2}{-0.5 "m/s"^2})) = 56.3^"o"$

The angle lies in the third quadrant since both the $x$ $x$ and the $y$ $y$ components of the vector are negative. The direction of the acceleration is ${56.3}^{o}$ $56.3^"o"$ counterclockwise from negative $x$ $x$ -axis.

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An object with a mass of $4$ $4$ $k g$ $kg$ is acted on by two forces. The first is $F_{1} = < - 4$ $F_1=<-4$ $N, 5$ $N, 5$ $N >$ $N>$ and the second is $F_{2} = < 2$ $F_2 = <2$ $N, - 8$ $N, -8$ $N >$ $N>$ . What is the object's rate and direction of acceleration?

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Explanation:

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An object with a mass of 444 kgkgkg is acted on by two forces. The first is F_1=<-4F1=<−4F_1=<-4 N, 5N,5N, 5 N>N>N> and the second is F_2 = <2F2=<2F_2 = <2 N, -8N,−8N, -8 N>N>N>. What is the object's rate and direction of acceleration?

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Explanation:

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An object with a mass of $4$ $4$ $k g$ $kg$ is acted on by two forces. The first is $F_{1} = < - 4$ $F_1=<-4$ $N, 5$ $N, 5$ $N >$ $N>$ and the second is $F_{2} = < 2$ $F_2 = <2$ $N, - 8$ $N, -8$ $N >$ $N>$ . What is the object's rate and direction of acceleration?