{"id":1045,"date":"2018-02-24T06:34:02","date_gmt":"2018-02-23T22:34:02","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1045"},"modified":"2018-02-24T06:34:02","modified_gmt":"2018-02-23T22:34:02","slug":"application-of-conservation-of-mechanical-energy-for-free-fall-motion","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/application-of-conservation-of-mechanical-energy-for-free-fall-motion.htm","title":{"rendered":"Application of conservation of mechanical energy for free fall motion – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. <\/span><\/span>A 1-kg body falls freely from rest, from a height of 80 m. Acceleration due to gravity<\/a> is <\/span><\/span>10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>. What is the kinetic energy<\/a> when the body hits the ground.<\/span><\/span><\/p>\n

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

Mass<\/a> (m) = 1 kg<\/span><\/span><\/p>\n

Height <\/span><\/span>(h) = 80 m<\/span><\/span><\/p>\n

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

Wanted:<\/u><\/span><\/span> kinetic energy when the body hits the ground<\/span><\/span><\/p>\n

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

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

M<\/span><\/span>E<\/span><\/span>o<\/span><\/sub> = P<\/span><\/span>E<\/span><\/span> = m g h = (1)(10)(80) = 800 Joule<\/span><\/span><\/p>\n

The final mechanical energy <\/span><\/span>(M<\/span><\/span>E<\/span><\/span>t<\/span><\/sub>) = <\/span><\/span>kinetic energy <\/span><\/span>(K<\/span><\/span>E<\/span><\/span>) <\/span><\/span><\/p>\n

The principle of conservation of mechanical energy<\/a> :<\/span><\/span><\/p>\n

M<\/span><\/span>E<\/span><\/span>o<\/span><\/sub> = M<\/span><\/span>E<\/span><\/span>t<\/span><\/sub><\/p>\n

P<\/span><\/span>E<\/span><\/span> = K<\/span><\/span>E<\/span><\/span><\/p>\n

800 = K<\/span><\/span>E<\/span><\/span><\/p>\n

The final kinetic energy is <\/span><\/span>800 Joule.<\/span><\/span><\/p>\n

<\/p>\n

2. <\/span><\/span>A 4-kg body free fall<\/a> from rest, from a height of 10 m. Acceleration due to gravity is <\/span><\/span>10 m s<\/span><\/span>\u20132<\/span><\/span><\/sup>. What is the kinetic energy and the velocity at 5 meters above the ground.<\/span><\/span><\/p>\n

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

The change in height <\/span><\/span>(h) = 10 \u2013 5 = 5 meter<\/span><\/span>s<\/span><\/span><\/p>\n

Mass (m) = 4 kg<\/span><\/span><\/p>\n

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

Wanted:<\/u> Kinetic energy at 5 meters above the ground and <\/span><\/span>the velocity at 5 meters above the ground<\/span><\/span><\/p>\n

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

(a) <\/u><\/span><\/span>Kinetic energy at 5 meters above the ground<\/u><\/span><\/span><\/p>\n

The initial mechanical energy <\/span><\/span>(M<\/span><\/span>E<\/span><\/span>o<\/span><\/sub>) = <\/span><\/span>the gravitational potential energy <\/span><\/span>(P<\/span><\/span>E<\/span><\/span>)<\/span><\/span><\/p>\n

M<\/span><\/span>E<\/span><\/span>o<\/span><\/sub> = P<\/span><\/span>E<\/span><\/span> = m g h = (4)(10)(5) = 200 Joule<\/span><\/span><\/p>\n

The final mechanical energy <\/span><\/span>(EM<\/span><\/span>t<\/span><\/sub>) = <\/span><\/span>kinetic energy <\/span><\/span>(EK)<\/span><\/span><\/p>\n

M<\/span><\/span>E<\/span><\/span>t<\/span><\/sub> = K<\/span><\/span>E<\/span><\/span><\/p>\n

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

M<\/span><\/span>E<\/span><\/span>o<\/span><\/sub> = M<\/span><\/span>E<\/span><\/span>t<\/span><\/sub><\/p>\n

200 = K<\/span><\/span>E<\/span><\/span><\/p>\n

Kinetic energy at 5 meters above the ground is <\/span><\/span>200 Joule.<\/span><\/span><\/p>\n

(b) <\/u><\/span><\/span>velocity at <\/u><\/span><\/span>5 meter<\/u><\/span><\/span>s above the ground<\/u><\/span><\/span><\/p>\n

The initial mechanical energy <\/span><\/span>(M<\/span><\/span>E<\/span><\/span>o<\/span><\/sub>) = <\/span><\/span>the final mechanical energy <\/span><\/span>(M<\/span><\/span>E<\/span><\/span>t<\/span><\/sub>)<\/span><\/span><\/p>\n

P<\/span><\/span>E<\/span><\/span> = K<\/span><\/span>E<\/span><\/span><\/p>\n

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

2(200) \/ 4 = v<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

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

v = \u221a100<\/span><\/span><\/p>\n

v = 10 <\/span><\/span>m\/s<\/span><\/span><\/p>\n

Body’s velocity at 5 meters above the ground is 10 m\/s. <\/span><\/span><\/p>\n

<\/p>\n

3. A mango <\/span><\/span>falls freely from rest, from a height of 2 meters. Acceleration due to gravity is <\/span><\/span>10 m s<\/span><\/span>\u20132<\/span><\/span><\/sup>. Determine mango’s velocity when hits the ground.<\/span><\/span><\/p>\n

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

Height <\/span><\/span>(h) = 2 meter<\/span><\/span>s<\/span><\/span><\/p>\n

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

Wanted :<\/u><\/span><\/span> mango’s velocity when hits the ground.<\/span><\/span><\/p>\n

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

The initial mechanical energy <\/span><\/span>(M<\/span><\/span>E<\/span><\/span>o<\/span><\/sub>) = <\/span><\/span>the gravitational potential energy <\/span><\/span>(P<\/span><\/span>E<\/span><\/span>)<\/span><\/span><\/p>\n

M<\/span><\/span>E <\/span><\/span>= P<\/span><\/span>E<\/span><\/span> = m g h = m (10)(2) = 20 m<\/span><\/span><\/p>\n

The final mechanical energy <\/span><\/span>(M<\/span><\/span>E<\/span><\/span>t<\/span><\/sub>) = <\/span><\/span>the kinetic energy <\/span><\/span>(K<\/span><\/span>E<\/span><\/span>)<\/span><\/span><\/p>\n

M<\/span><\/span>E<\/span><\/span>t<\/span><\/sub> = K<\/span><\/span>E<\/span><\/span> = \u00bd m v<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

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

M<\/span><\/span>E<\/span><\/span>o<\/span><\/sub> = M<\/span><\/span>E<\/span><\/span>t<\/span><\/sub><\/p>\n

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

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

2(20) = v<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

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

v = \u221a40 = \u221a(4)(10) = 2\u221a10 m\/<\/span><\/span>s<\/span><\/span><\/p>\n

[wpdm_package id=’1166′]<\/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 1-kg body falls freely from rest, from a height of 80 m. Acceleration due to gravity is 10 m\/s2. What is the kinetic energy when the body hits the ground. Known : Mass (m) = 1 kg Height (h) = 80 m Acceleration due to gravity (g) = 10 m\/s2 Wanted: kinetic 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 free fall motion - 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-1045","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\/1045","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=1045"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1045\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1045"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1045"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1045"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}