{"id":245,"date":"2023-06-23T10:06:26","date_gmt":"2023-06-23T10:06:26","guid":{"rendered":"https:\/\/physigeek.com\/pt\/forca-elastica-ou-forca-restauradora\/"},"modified":"2023-06-23T10:06:26","modified_gmt":"2023-06-23T10:06:26","slug":"forca-elastica-ou-forca-restauradora","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/forca-elastica-ou-forca-restauradora\/","title":{"rendered":"For\u00e7a el\u00e1stica (ou for\u00e7a restauradora)"},"content":{"rendered":"<p>Este artigo explica o que \u00e9 for\u00e7a el\u00e1stica (ou for\u00e7a restauradora). Assim, voc\u00ea descobrir\u00e1 como calcular a for\u00e7a el\u00e1stica, suas caracter\u00edsticas e os exerc\u00edcios de for\u00e7a el\u00e1stica resolvidos. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-la-fuerza-elastica\"><\/span> O que \u00e9 for\u00e7a el\u00e1stica?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>A for\u00e7a el\u00e1stica<\/strong> , tamb\u00e9m chamada de <strong>for\u00e7a restauradora<\/strong> , \u00e9 uma for\u00e7a exercida por um material el\u00e1stico quando ele se deforma. Mais precisamente, a for\u00e7a el\u00e1stica tem a mesma magnitude e dire\u00e7\u00e3o que a for\u00e7a que deformou o corpo el\u00e1stico, mas sua dire\u00e7\u00e3o \u00e9 oposta.<\/p>\n<p> Al\u00e9m disso, quanto maior for a deforma\u00e7\u00e3o sofrida pelo corpo el\u00e1stico, isto \u00e9, quanto mais o corpo el\u00e1stico tiver sido alongado ou comprimido, maior ser\u00e1 o m\u00f3dulo da for\u00e7a el\u00e1stica. <\/p>\n<figure class=\"wp-block-image aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-elastique.png\" alt=\"for\u00e7a el\u00e1stica\" class=\"wp-image-4161\" width=\"640\" height=\"318\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-elastique-300x150.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-elastique-1024x511.png 1024w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-elastique-768x383.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-elastique-1536x767.png 1536w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-elastique.png 1585w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p> Assim, uma mola sempre exerce a for\u00e7a el\u00e1stica na dire\u00e7\u00e3o oposta \u00e0 for\u00e7a externa aplicada a ela.<\/p>\n<p> Na f\u00edsica, problemas relacionados a molas s\u00e3o frequentemente abordados para compreender a no\u00e7\u00e3o de for\u00e7a el\u00e1stica. Veremos ent\u00e3o como a for\u00e7a el\u00e1stica \u00e9 calculada e como resolver problemas como este. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Formula-de-la-fuerza-elastica\"><\/span>F\u00f3rmula de for\u00e7a el\u00e1stica<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> A for\u00e7a el\u00e1stica exercida por uma mola \u00e9 igual a menos a constante el\u00e1stica da mola multiplicada pelo seu deslocamento.<\/p>\n<p> A <strong>f\u00f3rmula para a for\u00e7a el\u00e1stica<\/strong> \u00e9, portanto, a seguinte:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b886155aaf78dae4ad0a70c38f91d77f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_e=-k\\cdot \\Delta x\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"104\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p style=\"margin-bottom:5px\"> Ouro: <\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:8px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-88df03c55e081c7cd9da4e7d74ba7265_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"14\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 a for\u00e7a el\u00e1stica, expressa em newtons. <\/span><\/li>\n<li style=\"margin-bottom:8px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-d42bc2203d6f76ad01b27ac9acc0bee1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"k\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 a constante el\u00e1stica da mola, cujas unidades s\u00e3o N\/m.<\/span><\/li>\n<li><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-e936d3a449e3ecae93ebb5ae1e61feac_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\Delta x\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"25\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 o alongamento experimentado pela mola quando uma for\u00e7a externa \u00e9 aplicada, expresso em metros.<\/span><\/li>\n<\/ul>\n<p> <strong>Nota<\/strong> : O sinal negativo serve simplesmente para indicar que a dire\u00e7\u00e3o da for\u00e7a el\u00e1stica \u00e9 oposta \u00e0 for\u00e7a externa exercida na mola. O importante \u00e9 que o m\u00f3dulo da for\u00e7a el\u00e1stica seja equivalente \u00e0 constante el\u00e1stica multiplicada pelo deslocamento.<\/p>\n<p> Portanto, a f\u00f3rmula da for\u00e7a el\u00e1stica \u00e9 definida pela <a href=\"https:\/\/physigeek.com\/pt\/lei-de-hooke\/\">lei da elasticidade de Hooke<\/a> .<\/p>\n<p> Por outro lado, quando uma mola \u00e9 esticada ou comprimida, a energia potencial \u00e9 armazenada. Assim, a f\u00f3rmula para c\u00e1lculo da energia potencial el\u00e1stica \u00e9 a seguinte: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-3bd6ff61b959af0bf6c8061037ba754e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"E_p=\\cfrac{1}{2}\\cdot k \\cdot \\Delta x^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"123\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-de-la-fuerza-elastica\"><\/span>Exemplo de for\u00e7a el\u00e1stica<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Assim que vermos a defini\u00e7\u00e3o de for\u00e7a el\u00e1stica, veremos um exemplo resolvido de como esse tipo de for\u00e7a \u00e9 calculado.<\/p>\n<ul>\n<li> Uma mola com constante el\u00e1stica de 170 N\/m \u00e9 esticada ao longo de 45 cm. Qual \u00e9 a for\u00e7a el\u00e1stica que a mola exercer\u00e1?<\/li>\n<\/ul>\n<p> Para determinar a for\u00e7a el\u00e1stica, devemos utilizar a f\u00f3rmula que vimos acima:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b886155aaf78dae4ad0a70c38f91d77f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_e=-k\\cdot \\Delta x\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"104\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p> No entanto, antes de usar a f\u00f3rmula, voc\u00ea precisa converter o comprimento do deslocamento em metros:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-311621c96c01a4535b15801b7164ee17_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"45 \\ cm \\div 100 =0,45 \\ m\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"174\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p> Finalmente, substitu\u00edmos os dados da constante el\u00e1stica e do deslocamento da mola na f\u00f3rmula e calculamos a for\u00e7a el\u00e1stica: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-f6aa371674edfdd6c42d7e3bcb64383e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_e=-170\\cdot 0,45=-76,5 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"224\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-fuerza-elastica\"><\/span> Exerc\u00edcios resolvidos sobre for\u00e7a el\u00e1stica<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 1<\/h3>\n<p> Um objeto com massa de 8 kg est\u00e1 suspenso por uma mola vertical. Quanto a mola se esticar\u00e1 se sua constante el\u00e1stica for 350 N\/m? (g=10 m\/s <sup>2<\/sup> ) <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exemple-de-la-loi-de-hooke.png\" alt=\"exemplo resolvido da lei de Hooke\" class=\"wp-image-4066\" width=\"118\" height=\"229\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exemple-de-la-loi-de-hooke-154x300.png 154w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exemple-de-la-loi-de-hooke.png 283w\" sizes=\"auto, (max-width: 154px) 100vw, 154px\"><\/figure>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#FFF8E1\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Primeiro, precisamos calcular a for\u00e7a do peso exercido pela massa sobre a mola. Para fazer isso, basta multiplicar a massa pela gravidade:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-21b4daa2a5826fe5c1bffc3d93cb084a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=m\\cdot g = 8\\cdot 10=80 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"201\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E uma vez que conhecemos a for\u00e7a aplicada \u00e0 mola, podemos usar a f\u00f3rmula da for\u00e7a el\u00e1stica:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-d289d95f4fbd3eaaa673f0217091347c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_e=k\\cdot \\Delta x\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"90\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Resolvemos a extens\u00e3o da f\u00f3rmula:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-dd49ebc50db3810f4c2d06bf3716663d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\Delta x=\\cfrac{F_e}{k}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"69\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Por fim, substitu\u00edmos os valores na f\u00f3rmula e calculamos o alongamento da mola: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-82d67ff2ea6c812a73f07308c0600014_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\Delta x=\\cfrac{F_e}{k}=\\cfrac{80}{350} =0,23 \\ m = 23 \\ cm\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"271\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Exerc\u00edcio 2<\/h3>\n<p> Quando uma for\u00e7a de 50 N \u00e9 aplicada a uma mola, ela se estica 12 cm. Quanto a mola se esticar\u00e1 se uma for\u00e7a de 78 N for aplicada a ela? <\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#FFF8E1\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Para calcular o alongamento da mola, devemos primeiro determinar a sua constante el\u00e1stica. Portanto, resolvemos a constante el\u00e1stica a partir da f\u00f3rmula da for\u00e7a el\u00e1stica: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c0583a84bd59386c9674aaa94aa7d6ba_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F=k\\cdot \\Delta x \\quad \\longrightarrow \\quad k=\\cfrac{F}{\\Delta x}=\\cfrac{50}{0.12} =416.67 \\ \\cfrac{N} {m}[ \/latex] Maintenant que nous connaissons la valeur de la constante d'\u00e9lasticit\u00e9, nous pouvons calculer l'allongement du ressort en utilisant la loi de Hooke : [latex]F=k\\cdot \\Delta x \\quad \\longrightarrow \\quad \\Delta x=\\cfrac{F}{k}=\\cfrac{78}{416.67} =0,19 \\ m = 19 \\ cm \" title=\"Rendered by QuickLaTeX.com\" height=\"124\" width=\"1208\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Exerc\u00edcio 3<\/h3>\n<p> Temos uma bola de massa m = 7 kg colocada junto a uma mola em posi\u00e7\u00e3o horizontal cuja constante de elasticidade \u00e9 560 N\/m. Se empurrarmos a bola e comprimirmos a mola em 8 cm, ela empurra a bola e retorna \u00e0 sua posi\u00e7\u00e3o original. Com que acelera\u00e7\u00e3o a bola sair\u00e1 do contato com a mola? Despreze o atrito durante todo o exerc\u00edcio. <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-loi-du-crochet.png\" alt=\"exerc\u00edcio resoluto da lei de Hooke\" class=\"wp-image-4074\" width=\"295\" height=\"328\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-loi-du-crochet-270x300.png 270w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-loi-du-crochet.png 745w\" sizes=\"auto, (max-width: 270px) 100vw, 270px\"><\/figure>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#FFF8E1\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Primeiro, devemos calcular a for\u00e7a exercida ao empurrar a bola e comprimir a mola. Para fazer isso, aplicamos a f\u00f3rmula da lei de Hooke:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-02824e33202a9d71a6964aba33e74849_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F=k\\cdot \\Delta x=560 \\cdot 0,08 = 44,8 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"263\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para entender bem esta parte, voc\u00ea precisa ter clareza sobre o conceito de for\u00e7a el\u00e1stica. Quando uma for\u00e7a \u00e9 exercida sobre a mola, ela tamb\u00e9m produz uma for\u00e7a de rea\u00e7\u00e3o que tem a mesma magnitude e dire\u00e7\u00e3o, mas na dire\u00e7\u00e3o oposta (princ\u00edpio de a\u00e7\u00e3o-rea\u00e7\u00e3o). Assim, a for\u00e7a exercida pela mola sobre a bola tem o mesmo m\u00f3dulo que a for\u00e7a calculada acima:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-6b258ef25c5afc79b6401649152de803_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"|F_{ressort\\\u00e0 balle}|=|F|=44,8 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"219\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Finalmente, para determinar a acelera\u00e7\u00e3o da bola, devemos aplicar a segunda lei de Newton:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-3121171800c187fa5cf1a980aa5e80cc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_{spring\\to ball}=m_{ball}\\cdot a_{ball}\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"193\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ent\u00e3o resolvemos a acelera\u00e7\u00e3o da f\u00f3rmula e substitu\u00edmos os dados para encontrar o valor da acelera\u00e7\u00e3o da bola: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-f4dfeca33dce4b76e142d1a6e41f26f3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\" a_{ball}=\\cfrac{F_{spring\\to ball}}{m_{ball}}=\\cfrac{44,8}{7}=6,4 \\ \\cfrac{m}{s^2 }\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"279\" style=\"vertical-align: -15px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Este artigo explica o que \u00e9 for\u00e7a el\u00e1stica (ou for\u00e7a restauradora). Assim, voc\u00ea descobrir\u00e1 como calcular a for\u00e7a el\u00e1stica, suas caracter\u00edsticas e os exerc\u00edcios de for\u00e7a el\u00e1stica resolvidos. O que \u00e9 for\u00e7a el\u00e1stica? A for\u00e7a el\u00e1stica , tamb\u00e9m chamada de for\u00e7a restauradora , \u00e9 uma for\u00e7a exercida por um material el\u00e1stico quando ele se deforma. &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/forca-elastica-ou-forca-restauradora\/\"> <span class=\"screen-reader-text\">For\u00e7a el\u00e1stica (ou for\u00e7a restauradora)<\/span> Leia mais &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"","footnotes":""},"categories":[5],"tags":[],"class_list":["post-245","post","type-post","status-publish","format-standard","hentry","category-dinamico"],"yoast_head":"<!-- This site is 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