Question

Considere sempre g = 10 m/s2 e despreze a resistência do ar.


1- Considere um bloco de massa 8 kg, subindo um plano inclinado sem atrito, com inclinação de 30º (sen 30º = 0,5), onde atuam o peso (força da gravidade), a força normal e uma força de intensidade F, paralela ao plano, como na figura:



Calcule o valor de F em 3 situações distintas, sempre com o bloco subindo:

a) Em movimento acelerado, com aceleração de módulo 1,5 m/s2.

b) Em movimento uniforme.

c) Em movimento retardado, com aceleração de módulo 2,0 m/s2.

Answer 1

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$\LARGE\green{\boxed{\rm~~~\red{ 1-A)}~\gray{F}~\pink{=}~\blue{ 52~[N] }~~~}}$

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$\LARGE\green{\boxed{\rm~~~\red{ 1-B)}~\gray{F}~\pink{=}~\blue{ 40~[N] }~~~}}$

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$\LARGE\green{\boxed{\rm~~~\red{ 1-C)}~\gray{F}~\pink{=}~\blue{ 24~[N] }~~~}}$

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EXPLICAÇÃO PASSO-A-PASSO______✍

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☺lá, Miguel, como tens passado nestes tempos de quarentena⁉ E os estudos à distância, como vão⁉ Espero que bem❗ Acompanhe a resolução abaixo. ✌

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$\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\line(1,1){7}}\put(0,4){\line(1,1){2}}\put(2,2){\line(-1,1){2}}\put(4,4){\line(-1,1){2}}\put(2,4){\vector(-1,1){2}}\put(2,4){\vector(0,-1){3}}\put(2,4){\vector(1,1){2}}\put(2.1,1.3){\LARGE \sf F_p }\put(0.1,6){\LARGE \sf F_n }\bezier(1,1)(1.6,0.7)(1.5,0)\put(0.6,0.1){\Large \sf 30^{\circ} }\put(3.4,6){\LARGE\sf F}\end{picture}$

$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$ ☹ )

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☔ Decompondo nossa força peso tendo como referência para a orientação da inclinação da rampa teremos

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$\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\line(1,1){7}}\put(0,4){\line(1,1){2}}\put(2,2){\line(-1,1){2}}\put(4,4){\line(-1,1){2}}\put(2,4){\vector(-1,1){2}}\put(2,4){\vector(0,-1){3}}\put(2,4){\vector(1,1){2}}\put(0.1,6){\LARGE \sf F_n }\bezier(1,1)(1.6,0.7)(1.5,0)\put(0.6,0.1){\Large \sf 30^{\circ} }\put(2,4){\vector(1,-1){1.5}}\put(3.5,2.5){\vector(-1,-1){1.5}}\put(2.8,1.1){\LARGE \sf F_{p_x} }\put(2.6,3.5){\LARGE \sf F_{p_y} }\put(6,0){\dashbox{0.1}(4,4){}}\put(8,2){\vector(-1,-1){1.5}}\put(8,2){\vector(1,-1){1.5}}\put(8,2){\vector(-1,1){1.5}}\put(8,2){\vector(1,1){1.5}}\put(9.4,3){x}\put(7,3.3){y}\put(3.4,6){\LARGE\sf F}\end{picture}$

$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$ ☹ )

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$\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\line(1,1){7}}\put(0,4){\line(1,1){2}}\put(2,2){\line(-1,1){2}}\put(4,4){\line(-1,1){2}}\put(2,4){\vector(0,-1){3}}\bezier(1,1)(1.6,0.7)(1.5,0)\put(0.6,0.1){\Large \sf 30^{\circ} }\put(2,4){\vector(1,-1){1.5}}\put(3.5,2.5){\vector(-1,-1){1.5}}\bezier(2,1.8)(2.4,1.8)(2.5,1.5)\put(0,0){\line(1,0){7}}\put(2,1){\line(1,0){4}}\bezier(2.7,1.7)(3.2,1.4)(3,1)\put(2.4,1.1){ \sf 30^{\circ} }\put(2.03,1.43){ \sf 60^{\circ} }\put(2.7,2){\LARGE \sf F_{p_x} }\put(2.6,3.5){\LARGE \sf F_{p_y} }\end{picture}$

$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly$ ☹ )

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➡ $\Large\blue{\text{ \sf sen(30) = \dfrac{F_{p_x}}{F_p} }}$

➡ $\Large\blue{\sf F_{p_x} = F_p \cdot sen(30) }$

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➡ $\Large\blue{\text{ \sf cos(30) = \dfrac{F_{p_y}}{F_p} }}$

➡ $\Large\blue{\sf F_{p_y} = F_p \cdot cos(30) }$

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☔ Assumindo que não há deslocamento no eixo y ( $\sf F_{res_y} = N - F_{p_y} = 0$ ) e que $\sf F = m \cdot a$ então teremos em x que

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➡ $\large\blue{\sf F_{res_x} = F - F_{p_x} }$

➡ $\large\blue{\sf m \cdot a_x = F - m \cdot a_g \cdot sen(30) }$

➡ $\large\blue{\sf 8 \cdot a_x = F - 8 \cdot 10 \cdot \dfrac{1}{2} }$

➡ $\large\blue{\sf 8 \cdot a_x = F - 40 }$

➡ $\large\blue{\sf  F = 8 \cdot a_x + 40 }$

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Ⓐ$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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$\LARGE\gray{\boxed{\sf\blue{ a = 1,5~[m/s] }}}$

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➡ $\large\blue{\sf F = 8 \cdot 1,5 + 40 }$

$\large\blue{\sf = 12 + 40 }$

$\large\blue{\sf = 52~[N] }$

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$\LARGE\green{\boxed{\rm~~~\red{ A)}~\gray{F}~\pink{=}~\blue{ 52~[N] }~~~}}$ ✅

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Ⓑ$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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$\LARGE\gray{\boxed{\sf\blue{ a = 0 }}}$

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➡ $\large\blue{\sf F = 8 \cdot 0 + 40 }$

$\large\blue{\sf = 0 + 40 }$

$\large\blue{\sf = 40~[N] }$

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$\LARGE\green{\boxed{\rm~~~\red{ B)}~\gray{F}~\pink{=}~\blue{ 40~[N] }~~~}}$ ✅

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Ⓒ$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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$\LARGE\gray{\boxed{\sf\blue{ a = -2~[m/s] }}}$

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➡ $\large\blue{\sf F = 8 \cdot (-2) + 40 }$

$\large\blue{\sf = -16 + 40 }$

$\large\blue{\sf = 24~[N] }$

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$\LARGE\green{\boxed{\rm~~~\red{ C)}~\gray{F}~\pink{=}~\blue{ 24~[N] }~~~}}$ ✅

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$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

☕ $\bf\Large\blue{Bons\ estudos.}$

($\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios}$) ☄

$\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }}\LaTeX$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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$\gray{"Absque~sudore~et~labore~nullum~opus~perfectum~est."}$

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