{"id":1066,"date":"2018-02-24T16:11:23","date_gmt":"2018-02-24T08:11:23","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1066"},"modified":"2018-02-24T16:11:23","modified_gmt":"2018-02-24T08:11:23","slug":"power-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/power-problems-and-solutions.htm","title":{"rendered":"Power \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Power is the rate at which work is done in a given period. Mathematically, power is the work\/time ratio. <\/span><\/span><\/p>\n

P = W\/t<\/span><\/span><\/p>\n

Description: P = power (Joule\/second = Watt), W = work (Joule), t = time interval (second)<\/i><\/span><\/span><\/p>\n

Based on this equation, it can be concluded that the greater the work rate, the greater the power. On the other hand, the smaller the work rate, the smaller the power. Work rate refers to the rate at which work is done.<\/span><\/span><\/p>\n

<\/p>\n

Power is a scalar. The SI unit of power is the Joule\/second. The Joule\/second = Watt (abbreviated as W), named so to pay homage to James Watt. The British imperial unit for power is foot-pound per second.<\/span><\/span><\/p>\n

This unit is too small for practical purposes, so the greater unit horsepower (abbreviated as hp) is used. One horsepower = 550 foot-pound per second = 764 Watt = \u00be kilowatt.<\/span><\/span><\/p>\n

The amount of work can also be expressed in power x time units, for instance, kilowatt-hour or kWh. One kWh refers to the work done at a constant rate of 1 kiloWatt for one hour.<\/span><\/span><\/p>\n

1. <\/span><\/span>A 50-kg <\/span><\/span><\/span>person runs <\/span><\/span><\/span>up the stairs 10 meters high <\/span><\/span><\/span>in <\/span><\/span><\/span>2 minutes. <\/span><\/span><\/span>A<\/span><\/span><\/span>cceleration <\/span><\/span><\/span>due to <\/span><\/span><\/span>gravity (g) is 10 m\/<\/span><\/span><\/span>s<\/span><\/span><\/span>2<\/span><\/span><\/span><\/sup>. <\/span><\/span><\/span>Determine the power<\/a>.<\/span><\/span><\/span><\/p>\n

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

Mass<\/a> (m) = 50 kg<\/span><\/span><\/p>\n

Height <\/span><\/span>(h) = 10 m<\/span><\/span>eters<\/span><\/span><\/p>\n

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

Time interval <\/span><\/span>(t) = 2 <\/span><\/span>minute <\/span><\/span>= 2 (60) = 120 <\/span><\/span>second<\/span><\/span>s<\/span><\/span><\/p>\n

Wanted<\/u><\/span><\/span> : <\/span><\/span>Power<\/span><\/span> (P)<\/span><\/span><\/p>\n

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

Formula of power :<\/span><\/span><\/p>\n

P = W \/ t<\/span><\/span><\/p>\n

P = <\/i><\/span><\/span>power<\/i><\/span><\/span>, W = <\/i><\/span><\/span>work<\/i><\/span><\/span>, t = <\/i><\/span><\/span>time<\/i><\/span><\/span><\/p>\n

Formula of Work :<\/span><\/span><\/p>\n

W = F s = w h = m g h<\/span><\/span><\/p>\n

W = <\/span><\/span>work<\/span><\/span><\/a>, F = <\/span><\/span>force<\/span><\/span>, w = <\/span><\/span>weight<\/span><\/span><\/a>, <\/span><\/span>d<\/span><\/span> = <\/span><\/span>displacement<\/span><\/span><\/a>, h = <\/span><\/span>height<\/span><\/span>, m = mass, g = <\/span><\/span>acceleration due to gravity<\/span><\/span><\/i><\/p>\n

W = m g h = (50)(10)(10) = 5000 Joule.<\/span><\/span><\/p>\n

P = W \/ t = 5000 \/ 120 = 41.7 Joule\/se<\/span><\/span>c<\/span><\/span>on<\/span><\/span>d<\/span><\/span>.<\/span><\/span><\/p>\n

<\/p>\n

2. <\/span><\/span>Calculate the power required of a 60-kg person who climbs a tree 5 meters high in 10 seconds. Acceleration due to gravity is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>.<\/span><\/span><\/p>\n

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

Mass (m) = 60 kg<\/span><\/span><\/p>\n

Height <\/span><\/span>(h) = 5 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

Time interval <\/span><\/span>(t) = 10 <\/span><\/span>seconds<\/span><\/span><\/p>\n

Wanted <\/u><\/span><\/span>:<\/u><\/span><\/span> Power<\/span><\/span><\/p>\n

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

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

W = m g h = (60)(10)(5) = 3000 Joule<\/span><\/span><\/p>\n

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

P = W \/ t = 3000 \/ 10 = 300 Joule\/se<\/span><\/span>cond.<\/span><\/span><\/p>\n

3. A rotary comedy with a power of 300 watts and period of 5 minutes rotates 5 rounds. The energy it uses is ….<\/span><\/span><\/p>\n

A. 15 kJ<\/span><\/span><\/p>\n

B. 75 kJ<\/span><\/span><\/p>\n

C. 90 kJ<\/span><\/span><\/p>\n

D. 450 kJ<\/span><\/span><\/p>\n

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

Power (P) = 300 Watt = 300 Joule\/second<\/span><\/span><\/p>\n

Period (T) = 5 minutes = 5 (60 seconds) = 300 seconds<\/span><\/span><\/p>\n

Number of rotation = 5<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> Energy used by the rotary comedy<\/span><\/span><\/p>\n

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

\"Power<\/p>\n

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

[wpdm_package id=’1190′]<\/p>\n

    \n
  1. Work was 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":"

    Power is the rate at which work is done in a given period. Mathematically, power is the work\/time ratio. P = W\/t Description: P = power (Joule\/second = Watt), W = work (Joule), t = time interval (second) Based on this equation, it can be concluded that the greater the work rate, the greater the … 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":"Power \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-1066","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\/1066","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=1066"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1066\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1066"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1066"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1066"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}