Question

009 (part 1 of 3) 10.0 points A 4.8 kg object is subjected to two forces, F(2.3 N) i (-2.6 N) j and F (4.9 N) +(-11.9 N) . The object is at rest at the origin at time t -0. What is the magnitude of the object's ac- celeration? Answer in units of m/s?. 010 (part 2 of 3) 10.0 points What is the magnitude of the velocity at t- 3.8 s ? Answer in units of m/s. 011 (part 3 of 3) 10.0 points What is the magnitude of the object's position at t 3.8 s ? Answer in units of m 7:25 P

Answer 1

Given $\vec{F_{1}}=(2.3i-2.6j)N$

$\vec{F_{2}}=(4.9i-11.9j)N$

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Net Force =mass *acceleration

$\vec{F_{1}}+\vec{F_{2}}=m\vec{a}$

$(2.3i-2.6j)N+(4.9i-11.9j)N=4.8kg*\vec{a}$

$(7.2i-14.5j)=4.8*\vec{a}$

$(7.2i-14.5j)/4.8=\vec{a}$

$\vec{a}=(1.5i-3.021j)m/s^{2}$

$a=\sqrt{1.5^{2}+(-3.021)^{2}}$

ANSWER: ${\color{Red} a=3.373m/s^{2}}$

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Part 2

initially object is at rest

Use Formula $v=u+at$

$v=0m/s+3.373m/s^{2}*3.8s$

$v=0+3.373*3.8$

ANSWER: ${\color{Red} v=12.8174m/s}$

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Part 3

Use Formula $s=ut+1/2at^{2}$

initially object is at rest

$s=1/2at^{2}$

$s=0.5*3.373m/s^{2}*(3.8s)^{2}$

$s=0.5*3.373*3.8^{2}$

ANSWER: ${\color{Red} s=24.35m}$

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