{"id":241,"date":"2023-06-23T11:50:51","date_gmt":"2023-06-23T11:50:51","guid":{"rendered":"https:\/\/physigeek.com\/id\/hukum-hooke\/"},"modified":"2023-06-23T11:50:51","modified_gmt":"2023-06-23T11:50:51","slug":"hukum-hooke","status":"publish","type":"post","link":"https:\/\/physigeek.com\/id\/hukum-hooke\/","title":{"rendered":"Hukum hooke"},"content":{"rendered":"<p>Pada artikel ini Anda akan menemukan apa saja isi hukum Hooke, apa rumusnya, dan beberapa latihan yang diselesaikan langkah demi langkah pada hukum Hooke. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-la-ley-de-Hooke\"><\/span> Apa Hukum Hooke?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>Hukum Hooke<\/strong> , juga disebut <strong>hukum elastisitas Hooke<\/strong> , adalah hukum fisika yang menghubungkan gaya yang diterapkan pada pegas dengan perpanjangan pegas. Lebih spesifiknya, hukum Hooke menyatakan bahwa perpanjangan pegas berbanding lurus dengan besarnya gaya yang diberikan.<\/p>\n<p> Hukum Hooke ditemukan oleh fisikawan Inggris Robert Hooke. Menariknya, karena takut orang lain akan mempublikasikan penemuannya terlebih dahulu, Hooke pertama kali menerbitkan undang-undang tersebut dalam bentuk anagram pada tahun 1676, dan kemudian pada tahun 1678 ia menerbitkan undang-undang tersebut secara resmi.<\/p>\n<p> Hukum Hooke mempunyai banyak penerapan, dalam bidang teknik, konstruksi dan studi material, hukum Hooke banyak digunakan. Misalnya, pengoperasian dinamometer didasarkan pada hukum Hooke. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Formula-de-la-ley-de-Hooke\"><\/span> Rumus Hukum Hooke<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Hukum Hooke menyatakan bahwa gaya yang diterapkan pada pegas dan perpanjangannya berbanding lurus.<\/p>\n<p> Jadi, <strong>rumus hukum Hooke<\/strong> menyatakan bahwa gaya yang diterapkan pada pegas sama dengan hasil kali konstanta elastis pegas dan perpanjangannya.<\/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-7059f0a7fb501304d4de6360520575d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F=k\\cdot\\Delta x\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"85\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p style=\"margin-bottom:5px\"> Emas: <\/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> adalah gaya yang diterapkan pada pegas, dinyatakan dalam newton. <\/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> adalah konstanta elastis pegas yang satuannya 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> adalah perpanjangan yang dialami pegas ketika gaya diterapkan, dinyatakan dalam meter.<\/span><\/li>\n<\/ul>\n<p> Perlu diingat bahwa hukum Hooke hanya berlaku pada daerah elastis pegas, artinya ketika gaya berhenti maka pegas akan kembali ke bentuk semula. <\/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\/la-loi-de-hooke.png\" alt=\"hukum Hooke\" class=\"wp-image-4058\" width=\"464\" height=\"474\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/la-loi-de-hooke-293x300.png 293w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/la-loi-de-hooke-1002x1024.png 1002w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/la-loi-de-hooke-768x785.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/la-loi-de-hooke.png 1475w\" sizes=\"(max-width: 293px) 100vw, 293px\"><\/figure>\n<p> Ketika gaya luar diterapkan pada pegas, maka pegas akan menghasilkan gaya reaksi yang besar dan arahnya sama tetapi arahnya berlawanan (prinsip aksi-reaksi). Oleh karena itu, pegas akan selalu mengerahkan gaya untuk berusaha kembali ke posisi setimbangnya.<\/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-f90e909ce2ba85406d2bf43cb0922b17_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_{spring}=-k\\cdot \\Delta x\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"137\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p> Sebaliknya, dengan memberikan gaya pada pegas, energi potensial disimpan. Jadi rumus menghitung energi potensial elastis adalah: <\/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-23275e7a75b34506284c2ecb992e2844_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"U=\\cfrac{1}{2}\\cdot k \\cdot \\Delta x^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"115\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-de-la-ley-de-Hooke\"><\/span> Contoh Hukum Hooke<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Setelah kita mengetahui definisi hukum Hooke, berikut adalah contoh konkrit hukum fisika tersebut untuk memahami konsep tersebut secara utuh.<\/p>\n<ul>\n<li> Sebuah gaya sebesar 30 N diberikan pada sebuah pegas dan pegas tersebut memanjang sejauh 0,15 m. Berapakah konstanta elastis pegas tersebut?<\/li>\n<\/ul>\n<p> Dalam hal ini, ini adalah soal hukum Hooke karena kita mempelajari perpanjangan pegas, oleh karena itu kita harus menggunakan rumus yang terlihat di atas:<\/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-7059f0a7fb501304d4de6360520575d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F=k\\cdot\\Delta x\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"85\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Sekarang kita hilangkan konstanta elastisitas pegas dari rumus:<\/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-e23986b61dc1996f4d3bce1279bd11d7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"k=\\cfrac{F}{\\Delta x}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"60\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> Dan terakhir, kita substitusikan data masalah ke dalam rumus dan lakukan perhitungan: <\/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-162a0e44c922480a2815fea3a80e76ba_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"k=\\cfrac{F}{\\Delta x}=\\cfrac{30}{0.15}=200 \\ \\cfrac{N}{m}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"192\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-ley-de-Hooke\"><\/span> Memecahkan masalah hukum Hooke<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Latihan 1<\/h3>\n<p> Sebuah benda bermassa 8 kg digantung pada sebuah pegas vertikal. Berapa panjang pegas yang diperpanjang jika konstanta elastisnya 350 N\/m? (g=10m\/ <sup>s2<\/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=\"Contoh konkrit dari hukum 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=\"(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>lihat solusinya<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Pertama-tama, kita harus menghitung gaya beban yang diberikan massa pada pegas. Untuk melakukannya, cukup kalikan massa dengan gravitasi:<\/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\"> Dan setelah kita mengetahui gaya yang diterapkan pada pegas, kita dapat menggunakan rumus hukum 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-7059f0a7fb501304d4de6360520575d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F=k\\cdot\\Delta x\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"85\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Kami menghapus ekstensi dari rumus:<\/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-accbd15bbecc5a9bc18444e2f79bb5bc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\Delta x=\\cfrac{F}{k}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"64\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Terakhir, kami mengganti nilainya ke dalam rumus dan menghitung perpanjangan pegas: <\/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-60fd28dc6582cf40cce90571aa87a066_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\Delta x=\\cfrac{F}{k}=\\cfrac{80}{350} =0,23 \\ m = 23 \\ cm\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"266\" 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\">Latihan 2<\/h3>\n<p> Sebuah pegas diberi gaya sebesar 50 N, sehingga pegas tersebut memanjang sebesar 12 cm. Berapa panjang pegas jika diberi gaya sebesar 78 N? <\/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>lihat solusinya<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Untuk menghitung perpanjangan pegas, pertama-tama kita harus menentukan konstanta elastisnya. Oleh karena itu, kami mengisolasi konstanta pegas dari hukum Hooke dan menghitung nilainya: <\/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\">Latihan 3<\/h3>\n<p> Kita mempunyai sebuah bola bermassa m=7 kg yang diletakkan di sebelah pegas dengan posisi mendatar yang konstanta elastisitasnya 560 N\/m. Jika kita mendorong bola dan menekan pegas sebesar 8 cm, maka bola akan terdorong dan kembali ke kedudukan semula. Dengan percepatan berapa bola akan meninggalkan kontak dengan pegas? Abaikan gesekan selama latihan. <\/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=\"pelaksanaan hukum Hooke yang tegas\" 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=\"(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>lihat solusinya<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Pertama, kita harus menghitung gaya yang dilakukan dengan mendorong bola dan menekan pegas. Untuk melakukan ini, kami menerapkan rumus dari hukum 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\"> Untuk memahami bagian ini dengan baik, Anda harus memahami konsep hukum Hooke dengan jelas. Gaya yang diberikan pada pegas juga menghasilkan gaya reaksi yang besar dan arahnya sama, tetapi arahnya berlawanan. Jadi, gaya yang dikerjakan pegas pada bola sama besarnya dengan gaya yang dihitung di atas:<\/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\"> Terakhir, untuk menentukan percepatan bola, kita harus menerapkan hukum kedua 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\"> Jadi kita selesaikan percepatan dari rumus dan substitusikan datanya untuk mencari nilai percepatan bola:<\/p>\n<p class=\"has-text-align-center\"> [lateks] a_{bola}=\\cfrac{F_{pegas\\ke bola}}{m_{bola}}=\\cfrac{44.8}{7}=6.4 \\ \\cfrac{m}{s^2 }[\/lateks ]<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Pada artikel ini Anda akan menemukan apa saja isi hukum Hooke, apa rumusnya, dan beberapa latihan yang diselesaikan langkah demi langkah pada hukum Hooke. Apa Hukum Hooke? Hukum Hooke , juga disebut hukum elastisitas Hooke , adalah hukum fisika yang menghubungkan gaya yang diterapkan pada pegas dengan perpanjangan pegas. Lebih spesifiknya, hukum Hooke menyatakan bahwa &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/id\/hukum-hooke\/\"> <span class=\"screen-reader-text\">Hukum hooke<\/span> Baca selengkapnya &quot;<\/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":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>\u25b7 Hukum 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