{"id":1036,"date":"2018-02-23T09:07:35","date_gmt":"2018-02-23T01:07:35","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1036"},"modified":"2018-02-23T09:07:35","modified_gmt":"2018-02-23T01:07:35","slug":"potential-energy-of-elastic-spring-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/potential-energy-of-elastic-spring-problems-and-solutions.htm","title":{"rendered":"Potential energy of elastic spring – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. A 2-kg mass is attached to a spring. If the elongation of spring is 4 cm, determine potential energy of elastic spring<\/a>. Acceleration due to gravity<\/a> is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Mass<\/a> (m) = 2 kg<\/span><\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Weight<\/a> (w) = m g = (2)(10) = 20 N<\/span><\/span><\/p>\n

Elongation (x) = 4 cm = 0.04 m<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Potential energy of elastic spring<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

Formula of Hooke’s law<\/a> :<\/span><\/span><\/p>\n

F = k x<\/span><\/span><\/p>\n

F = force, k = spring constant, x = the change in length of spring<\/i><\/span><\/span><\/p>\n

Spring constant :<\/span><\/span><\/p>\n

k = w \/ x = 20 \/ 0.04 = 500 N\/m<\/span><\/span><\/p>\n

Potential energy of elastic spring :<\/span><\/span><\/p>\n

PE = \u00bd k x<\/span><\/span>2<\/span><\/span><\/sup> = \u00bd (500)(0.04)<\/span><\/span>2<\/span><\/span><\/sup> = (250)(0.0016) = 0.4 Joule<\/span><\/span><\/p>\n

Alternative solution :<\/span><\/span><\/p>\n

PE = \u00bd k x<\/span><\/span>2<\/span><\/span><\/sup> = \u00bd (w \/ x) x<\/span><\/span>2<\/span><\/span><\/sup> = <\/span><\/span>\u00bd w x<\/b><\/span><\/span> = \u00bd m g x <\/span><\/span><\/p>\n

w = weight, m = mass, x = the change in length of spring<\/i><\/span><\/span><\/p>\n

PE = \u00bd (2)(10)(0.04) = (10)(0.04) = 0.4 Joule.<\/span><\/span><\/p>\n

<\/p>\n

2. The elongation of spring stretched by a force F = 50 N is 10 cm. What is the potential energy of elastic spring if the elongation of spring is 12 cm.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Elongation (x) = 10 cm = 0.1 m<\/span><\/span><\/p>\n

Force (F) = 50 N<\/span><\/span><\/p>\n

Wanted <\/u><\/span><\/span>: The potential energy of elastic spring<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

Spring constant :<\/span><\/span><\/p>\n

k = F \/ x = 50 \/ 0.1 = 500 N\/m<\/span><\/span><\/p>\n

The potential energy of elastic spring if the elongation of spring is 12 cm. :<\/span><\/span><\/p>\n

PE = \u00bd k x<\/span><\/span>2<\/span><\/span><\/sup> = \u00bd (500)(0.12)<\/span><\/span>2<\/span><\/span><\/sup> = (250)(0.0144) = 3.6 Joule.<\/span><\/span><\/p>\n

<\/p>\n

3. Graph of F vs x shown in figure below. What is the potential energy of elastic spring, if the elongation of spring is 10 cm.<\/span><\/span><\/p>\n

\"Potential<\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Force (F) = 5 N<\/span><\/span><\/p>\n

The elongation of spring (x) = 2 cm = 0.02 m<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> The potential energy of elastic spring if the elongation of spring is 0.1 m.<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

Spring constant :<\/span><\/span><\/p>\n

k = F \/ x = 5 \/ 0.02 = 250 N\/m<\/span><\/span><\/p>\n

The potential energy of elastic spring if the elongation of spring is 0.1 m :<\/span><\/span><\/p>\n

PE = \u00bd k x<\/span><\/span>2<\/span><\/span><\/sup> = \u00bd (250)(0.1)<\/span><\/span>2<\/span><\/span><\/sup> = (125)(0.01) = 1.25 Joule.<\/span><\/span><\/p>\n

4<\/span><\/span><\/span>. <\/span><\/span><\/span>An athlete jumps onto a spring instrument with a weight of 500 N, the spring shortens 4 cm. Determine the amount of the potential energy to force the athlete.<\/span><\/span><\/span><\/p>\n

A. 20 Joule<\/span><\/span><\/p>\n

B. 10 Joule<\/span><\/span><\/p>\n

C. 5 Joule<\/span><\/span><\/p>\n

D. 2 Joule<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Force (F) = 500 N<\/span><\/span><\/p>\n

The change in length of spring (\u0394x) = 4 cm = 0.04 m<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> The potential energy of spring<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

Calculate the elastic constant of a spring using the equation of Hooke’s law :<\/span><\/span><\/p>\n

k = F \/ \u0394x = 500 N \/ 0.04 m = 12500 N\/m<\/span><\/span><\/p>\n

The elastic potential energy of spring :<\/span><\/span><\/p>\n

PE = \u00bd k \u0394x<\/span><\/span>2 <\/span><\/span><\/sup>= \u00bd (12500)(0.04)<\/span><\/span>2<\/span><\/span><\/sup> = (6250)(0.0016) <\/span><\/span><\/p>\n

PE = 10 Joule <\/span><\/span><\/p>\n

The correct answer is B.<\/span><\/span><\/p>\n

5. A spring when <\/span>suspended with a mass of <\/span>500 grams, <\/span>its length <\/span>increases by 5 cm. <\/span>Determine its <\/span>constant and <\/span>its <\/span>potential energy.<\/span><\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Mass of object (m) = 500 gram = (500\/1000) kg = 5\/10 kg = 0.5 kg <\/span><\/span><\/p>\n

Acceleration (g) = 10 m\/s2<\/sup><\/span><\/span><\/p>\n

Object weight (w) = m g = (0.5 kg)(10 m\/s2<\/sup>) = 5 kg m\/s2<\/sup> = 5 Newton<\/span><\/span><\/p>\n

The change in length of spring (\u0394x) = 5 cm = 0.05 m<\/span><\/span><\/p>\n

Wanted :<\/u> spring’s constant (k) and potential energy (PE)<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

First, calculate spring’s constant using equation of Hooke’s law :<\/span><\/span><\/p>\n

k = w \/ \u0394x = 5 N \/ 0.05 m = 100 N\/m<\/span><\/span><\/p>\n

The potential elastic of spring :<\/span><\/span><\/p>\n

PE = \u00bd k \u0394x2 <\/sup>= \u00bd (100)(0.05)2<\/sup> = (50)(0.0025) <\/span><\/span><\/p>\n

PE = 0.125 Joule<\/span><\/span><\/p>\n

[wpdm_package id=’1179′]<\/p>\n

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    <\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

    1. A 2-kg mass is attached to a spring. If the elongation of spring is 4 cm, determine potential energy of elastic spring. Acceleration due to gravity is 10 m\/s2. Known : Mass (m) = 2 kg Acceleration due to gravity (g) = 10 m\/s2 Weight (w) = m g = (2)(10) = 20 N … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Potential energy of elastic spring - problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-1036","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1036","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=1036"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1036\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1036"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1036"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1036"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}