{"id":200,"date":"2018-01-19T07:57:28","date_gmt":"2018-01-18T23:57:28","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=200"},"modified":"2018-01-19T07:57:28","modified_gmt":"2018-01-18T23:57:28","slug":"motion-on-inclined-plane-without-friction-force-application-of-newtons-law-of-motion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/motion-on-inclined-plane-without-friction-force-application-of-newtons-law-of-motion-problems-and-solutions.htm","title":{"rendered":"Motion on the inclined plane without the friction force – application of Newton’s law of motion problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. Box’s mass<\/a> = 2 kg, acceleration due to gravity<\/a> = 9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>. Find (a) the net force which accelerates the box downward (b) magnitude of the box’s acceleration<\/a>.<\/span><\/span><\/p>\n

\"Motion<\/p>\n

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

\"Motion<\/p>\n

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

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

Acceleration due to gravity (g) = 9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Weight<\/a> (w) = m g = (2)(9.8) = 19.6 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>x<\/span><\/sub> = w sin 30 = (19.6)(0.5) = 9.8 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>y<\/span><\/sub> = w cos 30 = (19.6)(0.5\u221a3) = 9.8\u221a3 Newton<\/span><\/span><\/p>\n

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

(a) The <\/span><\/span><\/span>net for<\/span>ce which accelerates the box<\/u><\/span><\/span><\/p>\n

Inclined plane is smooth, so there is no friction force. The only force which acts on the object is w<\/span><\/span>x<\/span><\/span><\/sub>. <\/span><\/span><\/p>\n

\u2211F = w<\/span><\/span>x<\/span><\/sub><\/p>\n

\u2211F = 9.8 Newton<\/span><\/span><\/p>\n

(b) <\/span><\/span>magnitude of the acceleration<\/u><\/span><\/span><\/p>\n

\u2211F = m a<\/span><\/span><\/p>\n

9.8 = (2) a<\/span><\/span><\/p>\n

a = 9.8 \/ 2 <\/span><\/span><\/p>\n

a = 4.9 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Magnitude of the acceleration is 4.9 m\/s<\/span><\/span>2<\/span><\/span><\/sup>, direction of the acceleration is downward.<\/span><\/span><\/p>\n

<\/p>\n

2. Inclined plane<\/a> is smooth so there is no friction force<\/a>. Object’s mass is 3 kg, acceleration due to gravity is 9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>. Determine the magnitude of the force F if (a) object is at rest (b) object is moving downward with constant acceleration 2 m\/s<\/span><\/span>2<\/span><\/span><\/sup> (c) object is moving upward with a constant acceleration of 2 m\/s<\/span><\/span>2<\/span><\/span><\/sup>.<\/span><\/span><\/p>\n

\"Motion<\/p>\n

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

\"Motion<\/p>\n

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

Mass (m) = 3 kg<\/span><\/span><\/p>\n

Acceleration due to gravity (g) = 9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Weight (w) = m g = (3)(9.8) = 29.4 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>x<\/span><\/sub> = w sin 30 = (29.4)(0.5) = 14.7 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>y <\/span><\/sub>= w cos 30 = (29.4)(0.5\u221a3) = 14.7\u221a3 Newton<\/span><\/span><\/p>\n

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

(a) The magnitude of the force F if an object is at rest<\/span><\/span><\/p>\n

Newton’s first law<\/a> of motion states that if an object is at rest, the net force acts on the object is zero. <\/span><\/span><\/p>\n

\u2211F = 0<\/span><\/span><\/p>\n

F \u2013 w<\/span><\/span>x<\/span><\/sub> = 0<\/span><\/span><\/p>\n

F = w<\/span><\/span>x<\/span><\/sub><\/p>\n

F = 14.7 Newton<\/span><\/span><\/p>\n

(b) The magnitude of the force F if an object is moving downward at a constant 2 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

\u2211F = m a<\/span><\/span><\/p>\n

w<\/span><\/span>x<\/span><\/sub> \u2013 F = m a<\/span><\/span><\/p>\n

14.7 \u2013 F = (3)(2)<\/span><\/span><\/p>\n

14.7 \u2013 F = 6<\/span><\/span><\/p>\n

F = 14.7\u2013 6<\/span><\/span><\/p>\n

F = 8.7 Newton<\/span><\/span><\/p>\n

(c) The magnitude of the force F if an object is moving upward at a constant 2 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

\u2211F = m a<\/span><\/span><\/p>\n

F – w<\/span><\/span>x<\/span><\/sub> = m a<\/span><\/span><\/p>\n

F \u2013 14.7 = (3)(2)<\/span><\/span><\/p>\n

F \u2013 14.7 = 6<\/span><\/span><\/p>\n

F = 14.7 + 6<\/span><\/span><\/p>\n

F = 20.7 Newton<\/span><\/span><\/p>\n

<\/p>\n

[wpdm_package id=’479′]<\/p>\n

    \n
  1. Mass and weight<\/a><\/li>\n
  2. Normal force<\/a><\/li>\n
  3. Newton’s second law of motion<\/a><\/li>\n
  4. Friction force<\/a><\/li>\n
  5. Motion on the horizontal surface without friction force<\/a><\/li>\n
  6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force<\/a><\/li>\n
  7. Motion on the inclined plane without friction force<\/a><\/li>\n
  8. Motion on the rough inclined plane with the friction force<\/a><\/li>\n
  9. Motion in an elevator<\/a><\/li>\n
  10. The motion of bodies connected by cord and pulley<\/a><\/li>\n
  11. Two bodies with the same magnitude of accelerations<\/a><\/li>\n
  12. Rounding a flat curve – dynamics of circular motion<\/a><\/li>\n
  13. Rounding a banked curve – dynamics of circular motion<\/a><\/li>\n
  14. Uniform motion in a horizontal circle<\/a><\/li>\n
  15. Centripetal force in uniform circular motion<\/a><\/li>\n<\/ol>\n

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

    1. Box’s mass = 2 kg, acceleration due to gravity = 9.8 m\/s2. Find (a) the net force which accelerates the box downward (b) magnitude of the box’s acceleration. Solution Known : Mass (m) = 2 kg Acceleration due to gravity (g) = 9.8 m\/s2 Weight (w) = m g = (2)(9.8) = 19.6 Newton … 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":"Motion on the inclined plane without the friction force - application of Newton's law of 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-200","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\/200","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=200"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/200\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=200"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=200"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=200"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}