{"id":2113,"date":"2018-04-24T14:17:42","date_gmt":"2018-04-24T06:17:42","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2113"},"modified":"2023-08-09T04:02:18","modified_gmt":"2023-08-09T04:02:18","slug":"mechanical-energy-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/mechanical-energy-problems-and-solutions.htm","title":{"rendered":"Mechanical energy \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Mechanical energy \u2013 problems and solutions<\/span><\/p>\n

The work-mechanical energy principle<\/b><\/span><\/a><\/span><\/p>\n

1. The coefficient of the kinetic friction<\/a> between block and floor (\u03bck<\/sub>) is 0.5. What is the displacement<\/a> of an object (s)? Acceleration due to gravity<\/a> is 10 m\/s2<\/sup>.<\/span><\/p>\n

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

The coefficient of the kinetic friction (\u03bck<\/sub>) = 0.5<\/span><\/p>\n

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

Acceleration due to gravity (g) is 10 m\/s2<\/sup><\/span><\/p>\n

Weight<\/a> of block (w) = m g = (4)(10) = 40 Newton<\/span><\/p>\n

If block on the horizontal plane then the normal force (N) =weight (w) = 40 Newton.<\/span><\/p>\n

If block on the horizontal plane then the normal force (N) = weight (w) = 40 Newton. <\/span><\/p>\n

Initial velocity (v1<\/sub>) = 5 m\/s<\/span><\/p>\n

Final velocity (v2<\/sub>) = 0 m\/s<\/span><\/p>\n

Wanted :<\/u> displacement of object (d) ?<\/span><\/p>\n

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

The work-mechanical energy principle states that work (W) done by the nonconservative force is the same as the change of the mechanical energy of an object. The change of mechanical energy = the final mechanical energy \u2013 the initial mechanical energy.<\/span><\/p>\n

The kinetic friction force is one of the nonconservative forces and the only one nonconservative force that acts on the block. <\/span><\/p>\n

W = \u0394ME<\/span><\/p>\n

fk <\/sub>d = ME2<\/sub> \u2013 ME1<\/sub><\/span><\/p>\n

Work done by the kinetic friction force :<\/span><\/p>\n

W = fk<\/sub> d = (\u03bck<\/sub>)(N)(d) = (0.5)(40)(d) = 20 d<\/span><\/p>\n

The change of the mechanical energy :<\/span><\/p>\n

\u0394ME = ME2<\/sub> \u2013 ME1<\/sub> = (KE + PE)2<\/sub> \u2013 (KE + PE)1<\/sub><\/span><\/p>\n

Object moves along the horizontal plane and no change of height (\u0394h = 0) so there is no change of the gravitational potential energy<\/a> (\u0394PE = PE2<\/sub> \u2013 PE1<\/sub> = 0). Thus the change of the mechanical energy<\/a> just involves the change of the kinetic energy<\/a>.<\/span><\/p>\n

\u0394ME = KE2<\/sub> \u2013 KE1<\/sub> = \u00bd m v2<\/sub>2<\/sup> \u2013 \u00bd m v1<\/sub>2 <\/sup>= \u00bd m (v2<\/sub>2 <\/sup>– v1<\/sub>2<\/sup>)<\/span><\/p>\n

\u0394<\/b>M<\/b>E<\/b> = \u00bd (4)(0<\/b>2<\/b><\/sup> \u2013 5<\/b>2<\/b><\/sup>) = (2)(25) = 50<\/b><\/span><\/p>\n

Displacement of block :<\/u><\/span><\/p>\n

W = \u0394ME<\/span><\/p>\n

20 s = 50<\/span><\/p>\n

s = 50 \/ 20 <\/span><\/p>\n

s = 2.5 meters<\/span><\/p>\n

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

2. Object A and object B have the same mass. Object A free fall from a height of h meters and object B free fall from a height of 2h meters. If object A hits the ground at v m\/s, then what is the kinetic energy of the object B when it hits the ground.<\/span><\/p>\n

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

The final velocity of object B when free fall from a height of 2h :<\/span><\/p>\n

v2<\/sup> = 2 g (2h) = 4 g h<\/span><\/p>\n

The kinetic energy of object B :<\/span><\/p>\n

KEB<\/sub> = \u00bd m v2<\/sup> = \u00bd m (4 g h) = 2 m g h —– equation 1<\/span><\/p>\n

The initial mechanical energy of object B = the gravitational potential energy = m g h.<\/span><\/p>\n

The final mechanical energy of object B = the kinetic energy = \u00bd m v2<\/sup>. <\/span><\/p>\n

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

m g h = \u00bd m v2<\/sup><\/span><\/p>\n

Because m g h = \u00bd m v<\/u>2<\/u><\/sup> then we can change m g h<\/u> in equation 1 with \u00bd m v2<\/sup>. <\/span><\/p>\n

The kinetic energy of object B = 2 m g h = 2(\u00bd m v2<\/sup>) = m v2<\/sup><\/span><\/p>\n

3. An object free fall from a height of 20 meters. Acceleration due to gravity is 10 m\/s. What is the velocity of object 15 meters above the ground?<\/span><\/p>\n

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

The final mechanical energy = the initial mechanical energy<\/span><\/p>\n

The kinetic energy at point 2 = the change of the gravitational potential energy as far as 5 meters.<\/span><\/p>\n

\"Mechanical<\/span><\/p>\n

4. A block is released from the top of the smooth inclined plane. What is the velocity of the block when hits the ground?<\/span><\/p>\n

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

The initial mechanical energy = the gravitational potential energy = m g h = m (10)(5) = 50 meters\"Mechanical<\/span><\/p>\n

The final mechanical energy = the kinetic energy = 1\/2 m v2 <\/sup> <\/span><\/p>\n

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

MEo<\/sub> = MEt<\/sub><\/span><\/p>\n

50 m = 1\/2 m v2 <\/sup> <\/span><\/p>\n

50 = 1\/2 v2 <\/sup> <\/span><\/p>\n

2 (50) = v2 <\/sup> <\/span><\/p>\n

100 = v2 <\/sup> <\/span><\/p>\n

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

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

A block with mass of m-kg released from a height of h meters above the ground, as shown in figure below. Determine the ratio of the potential energy to the kinetic energy (KE) at point M.<\/span><\/p>\n

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

The gravitational potential energy at point M :\"Mechanical<\/span><\/p>\n

PEM<\/sub> = m g (0.3 h)<\/span><\/p>\n

The kinetic energy at point m = the change of the gravitational potential energy as far as h-0.3h = 0.7 h<\/span><\/p>\n

KEM<\/sub> = PE = m g (0.7 h)<\/span><\/p>\n

The ratio of the gravitational potential energy to the kinetic energy at point M :<\/span><\/p>\n

PEM<\/sub> : KEM<\/sub><\/span><\/p>\n

m g (0.3 h) : m g (0.7 h)<\/span><\/p>\n

0.3 : 0.7<\/span><\/p>\n

3 : 7<\/span><\/p>\n

6. If PEQ<\/sub> and KEQ<\/sub> have the potential energy and the kinetic energy at point Q (g = 10 m\/s2<\/sup>), then PEQ<\/sub> : KEQ<\/sub> =…<\/span><\/p>\n

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

The gravitational potential energy at point Q :\"Mechanical<\/span><\/p>\n

PEQ<\/sub> = m g h = (m)(10)(1.8) = 18 meters<\/span><\/p>\n

The kinetic energy at point Q = the change of the gravitational potential energy as far as 5-1.8 = 3.2 meters<\/span><\/p>\n

KEQ<\/sub> = PE = m g h = m (10)(3.2) = 32 m<\/span><\/p>\n

The ratio of the gravitational potential energy to the kinetic energy at point Q :<\/span><\/p>\n

PEQ<\/sub> : KEQ <\/sub><\/span><\/p>\n

18 m : 32 m<\/span><\/p>\n

18 : 32<\/span><\/p>\n

9 : 16<\/span><\/p>\n

    \n
  1. What is mechanical energy?<\/strong> Answer<\/em>: Mechanical energy is the sum of potential energy and kinetic energy in a system. It represents the total amount of energy associated with the motion and position of an object.<\/span><\/li>\n
  2. How is kinetic energy different from potential energy?<\/strong> Answer<\/em>: Kinetic energy is the energy of motion, while potential energy is the energy stored due to an object’s position or configuration. For example, a moving car has kinetic energy, while a stretched spring or an object at height in a gravitational field has potential energy.<\/span><\/li>\n
  3. Is mechanical energy always conserved?<\/strong> Answer<\/em>: In an ideal, closed system with no external forces, mechanical energy is conserved. However, in real-world situations, other forms of energy like thermal or sound energy can come into play due to friction or other non-conservative forces, leading to a decrease in total mechanical energy.<\/span><\/li>\n
  4. How does the conservation of mechanical energy help in solving problems?<\/strong> Answer<\/em>: When mechanical energy is conserved in a system (no non-conservative forces like friction), the total energy at the beginning is equal to the total energy at the end. This principle allows us to relate the potential and kinetic energies at different points in time, simplifying the problem-solving process.<\/span><\/li>\n
  5. What happens to the mechanical energy of a ball when it’s thrown upwards and comes to a momentary stop before descending?<\/strong> Answer<\/em>: As the ball rises, its kinetic energy decreases while its potential energy increases. At the peak, all its mechanical energy is potential energy. As it descends, this process reverses. In an ideal scenario without air resistance, the ball’s total mechanical energy remains constant throughout its journey.<\/span><\/li>\n
  6. Can an object have both kinetic and potential energy at the same time?<\/strong> Answer<\/em>: Yes, an object can possess both forms of energy simultaneously. For instance, a pendulum at the midpoint of its swing has kinetic energy due to its motion and potential energy due to its height above its lowest point.<\/span><\/li>\n
  7. How does a roller coaster demonstrate the conservation of mechanical energy?<\/strong> Answer<\/em>: A roller coaster starts with potential energy at its highest point. As it descends, this potential energy converts to kinetic energy, making the coaster speed up. As it climbs again, the kinetic energy reduces while potential energy increases. Ignoring friction and air resistance, the total mechanical energy remains nearly constant.<\/span><\/li>\n
  8. What factors can lead to a loss of mechanical energy in real-world systems?<\/strong> Answer<\/em>: Friction, air resistance, and sound generation are some factors that can lead to energy losses in real-world systems. The energy isn’t destroyed (due to the law of conservation of energy) but is transformed into other forms like heat or sound.<\/span><\/li>\n
  9. What is the relationship between work and mechanical energy?<\/strong> Answer<\/em>: Work done on a system by external forces can lead to a change in the system’s mechanical energy. In mathematical terms, work is equal to the change in mechanical energy.<\/span><\/li>\n
  10. \n

    If gravitational potential energy is relative to a reference point, how can we consistently measure changes in it?<\/strong> Answer<\/em>: While the absolute value of gravitational potential energy is indeed dependent on the choice of reference point, changes in potential energy (like when an object is raised or lowered) are consistent regardless of the reference. Thus, we often focus on changes in potential energy rather than absolute values.<\/span><\/p>\n<\/li>\n<\/ol>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

    Mechanical energy \u2013 problems and solutions The work-mechanical energy principle 1. The coefficient of the kinetic friction between block and floor (\u03bck) is 0.5. What is the displacement of an object (s)? Acceleration due to gravity is 10 m\/s2. Known : The coefficient of the kinetic friction (\u03bck) = 0.5 Mass of block (m) = … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"closed","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":"Mechanical energy \u2013 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-2113","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\/2113","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=2113"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2113\/revisions"}],"predecessor-version":[{"id":8645,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2113\/revisions\/8645"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=2113"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=2113"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=2113"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}