{"id":1061,"date":"2018-02-24T12:39:28","date_gmt":"2018-02-24T04:39:28","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1061"},"modified":"2018-02-24T12:39:28","modified_gmt":"2018-02-24T04:39:28","slug":"application-of-conservation-of-mechanical-energy-for-motion-on-inclined-plane","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/application-of-conservation-of-mechanical-energy-for-motion-on-inclined-plane.htm","title":{"rendered":"Application of conservation of mechanical energy for motion on inclined plane – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. A block slides down on smooth inclined plane<\/a> without friction<\/a>. What is block’s velocity<\/a> when hits the ground. Acceleration due to gravity is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

\"ApplicationKnown :<\/u><\/span><\/span><\/p>\n

Height (h) = 8 m<\/span><\/span><\/p>\n

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

Wanted<\/u><\/span><\/span> : velocity (v)<\/span><\/span><\/p>\n

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

Initial mechanical energy (ME<\/u><\/span><\/span>o<\/u><\/span><\/sub>) = gravitational potential energy<\/a> (PE)<\/u><\/span><\/span><\/p>\n

ME<\/span><\/span>o<\/span><\/sub> = PE = m g h = m (10)(8) = 80 m<\/span><\/span><\/p>\n

Final mechanical energy (ME<\/u><\/span><\/span>t<\/u><\/span><\/span><\/sub>) = kinetic energy<\/a> (KE)<\/u><\/span><\/span><\/p>\n

ME<\/span><\/span>t<\/span><\/sub> = KE = \u00bd m v<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Principle of conservation of mechanical energy states that the initial mechanical energy<\/a> = the final mechanical energy :<\/span><\/span><\/p>\n

ME<\/span><\/span>o<\/span><\/sub> = ME<\/span><\/span>t<\/span><\/sub><\/p>\n

80 m = \u00bd m v<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

80 = \u00bd v<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

160 = v<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

v = \u221a160 = \u221a(16)(10) = 4\u221a10 m\/s<\/span><\/span><\/p>\n

<\/p>\n

2. A 1-kg object slides down along 8 meters. Determine kinetic energy after the object moves along 5 meters… Acceleration due to gravity g = 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

\"ApplicationKnown :<\/u><\/span><\/span><\/p>\n

Mass (m) = 0.2 kg<\/span><\/span><\/p>\n

d = 5 meters<\/span><\/span><\/p>\n

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

Wanted<\/u><\/span><\/span> : kinetic energy (KE) <\/span><\/span><\/p>\n

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

sin 30<\/span><\/span>o<\/span><\/span><\/sup> = h \/ d<\/span><\/span><\/p>\n

0.5 = h \/ 5 <\/span><\/span><\/p>\n

h = (0.5)(5) = 2.5 meters<\/span><\/span><\/p>\n

The change in height of the object is 2.5 meters.<\/span><\/span><\/p>\n

The initial mechanical energy (ME<\/u><\/span><\/span>o<\/u><\/span><\/span><\/sub>) = the gravitational potential energy (PE)<\/u><\/span><\/span><\/p>\n

ME<\/span><\/span>o<\/span><\/sub> = PE = m g h = (1)(10)(2.5) = 25 Joule<\/span><\/span><\/p>\n

The final mechanical energy (ME<\/u><\/span><\/span>t<\/u><\/span><\/span><\/sub>) = kinetic energy (KE)<\/u><\/span><\/span><\/p>\n

ME<\/span><\/span>t<\/span><\/span><\/sub> = KE<\/span><\/span><\/p>\n

The principle of conservation of mechanical energy<\/a> states that the initial mechanical energy = the final mechanical energy :<\/span><\/span><\/p>\n

ME<\/span><\/span>o<\/span><\/sub> = ME<\/span><\/span>t<\/span><\/span><\/sub><\/p>\n

25 = KE<\/span><\/span><\/p>\n

Kinetic energy<\/a> = 25 Joule.<\/span><\/span><\/p>\n

[wpdm_package id=’1170′]<\/p>\n

    \n
  1. Work done by force problems and solutions<\/a><\/li>\n
  2. Work-kinetic energy problems and solutions<\/a><\/li>\n
  3. Work-mechanical energy principle problems and solutions<\/a><\/li>\n
  4. Gravitational potential energy problems and solutions<\/a><\/li>\n
  5. The potential energy of elastic spring problems and solutions<\/a><\/li>\n
  6. Power problems and solutions<\/a><\/li>\n
  7. Application of conservation of mechanical energy for free fall motion<\/a><\/li>\n
  8. Application of conservation of mechanical energy for up and down motion in free fall motion<\/a><\/li>\n
  9. Application of conservation of mechanical energy for motion on a curved surface<\/a><\/li>\n
  10. Application of conservation of mechanical energy for motion on an inclined plane<\/a><\/li>\n
  11. Application of conservation of mechanical energy for projectile motion<\/a><\/li>\n<\/ol>\n

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

    1. A block slides down on smooth inclined plane without friction. What is block’s velocity when hits the ground. Acceleration due to gravity is 10 m\/s2 Known : Height (h) = 8 m Acceleration due to gravity (g) = 10 m\/s2 Wanted : velocity (v) Solution : Initial mechanical energy (MEo) = gravitational potential energy … 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":"Application of conservation of mechanical energy for motion on inclined plane - 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-1061","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\/1061","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=1061"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1061\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1061"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1061"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1061"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}