{"id":4911,"date":"2018-11-04T23:48:46","date_gmt":"2018-11-05T07:48:46","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=4911"},"modified":"2018-11-04T23:48:46","modified_gmt":"2018-11-05T07:48:46","slug":"parabolic-motion-work-and-kinetic-energy-linear-momentum-linear-and-angular-motion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/parabolic-motion-work-and-kinetic-energy-linear-momentum-linear-and-angular-motion-problems-and-solutions.htm","title":{"rendered":"Parabolic motion Work and kinetic energy Linear momentum Linear and angular motion Problems and Solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

5 Parabolic motion Work and kinetic energy Linear momentum Linear and angular motion Problems and Solutions<\/span><\/h3>\n

1. A ball is thrown from the top of a building with an initial speed<\/a> of 8 m\/s at an angle of 20o<\/sup> below the horizontal. The ball hits the ground 3 seconds later.<\/span><\/span><\/p>\n

a) How far from the bottom of building the ball touches the ground<\/span><\/span><\/p>\n

b) How high is the ball thrown from the place <\/span><\/span><\/p>\n

c) How long the ball reaches a height of 10 m from the place of throwing<\/span><\/span><\/p>\n

\"Parabolicvox<\/sub> = vo<\/sub> cos 20o <\/sup>= 8 (0.939) = 7.512 m\/s<\/span><\/span><\/p>\n

voy<\/sub> = vo<\/sub> sin 20o <\/sup>= 8 (0.342) = 2.736 m\/s<\/span><\/span><\/p>\n

a) x<\/span><\/span><\/p>\n

Horizontal movements are analyzed as uniform linear motion,<\/a> so that used the formula for uniform linear motion.<\/span><\/span><\/p>\n

x = vox<\/sub> t<\/span><\/span><\/p>\n

x = (7.512 m\/s)(3 s) <\/span><\/span><\/p>\n

x = 22.536 m<\/span><\/span><\/p>\n

b) y<\/span><\/span><\/p>\n

The movements of the vertical direction are analyzed as the free fall motion<\/a> using the free fall motion formula.<\/span><\/span><\/p>\n

Use this formula y = vo<\/sub> t + \u00bd g t2<\/sup><\/span><\/span><\/p>\n

In free fall motion, vo<\/sub> = 0 so the formula becomes : y = \u00bd g t2<\/sup><\/span><\/span><\/p>\n

y = \u00bd (9.8)(32<\/sup>)<\/span><\/span><\/p>\n

y = (4.9)(9)<\/span><\/span><\/p>\n

y = 44 m<\/span><\/span><\/p>\n

c) y = 10 m<\/span><\/span><\/p>\n

Use the formula y = \u00bd g t2<\/sup><\/span><\/span><\/p>\n

10 = \u00bd (9,8) t2<\/sup><\/span><\/span><\/p>\n

10 = 4.9 t2<\/sup><\/span><\/span><\/p>\n

t2 <\/sup>= 10 \/ 4.9<\/span><\/span><\/p>\n

t2 <\/sup>= 2<\/span><\/span><\/p>\n

t = 1.4 s<\/span><\/span><\/p>\n

2. A dog train team pulls a mass of 100 kg load as far as 2 km above the horizontal surface at a constant speed. If the coefficient of friction between the train and snow is 0.15, determine:<\/span><\/span><\/p>\n

a) the work done by the dog<\/span><\/span><\/p>\n

\"ParabolicCalculate the kinetic friction<\/a>:<\/span><\/span><\/p>\n

Fk <\/sub>= \u00b5k<\/sub> N = \u00b5k<\/sub> w = \u00b5k<\/sub> m g = (0.15)(100 kg)(9.8 m\/s2<\/sup>) =147 N<\/span><\/span><\/p>\n

The speed of mass is constant so, according to Newton’s law, the resultant force = 0. So Fk <\/sub>= F = 147 N<\/span><\/span><\/p>\n

The work<\/a> done by the dog:<\/span><\/span><\/p>\n

W = F d = (147 N)(2000 m) = 294,000 J <\/span><\/span><\/p>\n

b) Energy lost by friction<\/span><\/span><\/p>\n

Energy lost by friction:<\/span><\/span><\/p>\n

Work = energy, so:<\/span><\/span><\/p>\n

W = Fk<\/sub> d = (147 N)(2000 m) = 294,000 J <\/span><\/span><\/p>\n

3. The 150 N horizontal force is used to push the box 6 meters above the rough surface. If the box moves at a constant speed:<\/span><\/span><\/p>\n

a) Work done by the force of 150 N<\/span><\/span><\/p>\n

W = F d = (150 N) (6 m) = 900 J<\/span><\/span><\/p>\n

b) Energy lost due to friction<\/span><\/span><\/p>\n

The box moves at a constant speed so that the resultant force acting on the box = 0.<\/span><\/span><\/p>\n

So, the horizontal force = 150 N = kinetic friction.<\/span><\/span><\/p>\n

Energy lost due to friction:<\/span><\/span><\/p>\n

W = Fk<\/sub> s = (150 N)(6 m) = 900 J<\/span><\/span><\/p>\n

c) The coefficient of kinetic friction <\/span><\/span><\/p>\n

The formula of friction force:<\/span><\/span><\/p>\n

Fk <\/sub>= \u00b5k<\/sub> N = \u00b5k<\/sub> w = \u00b5k<\/sub> m g<\/span><\/span><\/p>\n

150 = \u00b5k<\/sub> m (10)<\/span><\/span><\/p>\n

15 = \u00b5k<\/sub> m <\/span><\/span><\/p>\n

\u00b5k<\/sub> = 15\/m<\/span><\/span><\/p>\n

m = mass of the box<\/span><\/span><\/p>\n

\u00b5k<\/sub> = coefficient of kinetic friction <\/span><\/span><\/p>\n

Kinetic friction works when objects move, static friction works when objects are still stationary.<\/span><\/span><\/p>\n

4. A particle has a speed (3i-4j) m\/s. Determine the momentum component on the x and y-axis and the magnitude of momentum.<\/span><\/span><\/p>\n

Speed on the x-axis: 3 m\/s<\/span><\/span><\/p>\n

Speed on the y-axis: 4 m\/s<\/span><\/span><\/p>\n

Momentum on the x-axis: m v = 3 m<\/span><\/span><\/p>\n

momentum on the y-axis: m v = 4 m<\/span><\/span><\/p>\n

The magnitude of momentum:<\/span><\/span><\/p>\n

p2<\/sup> = (3 m)2 <\/sup>+ (4 m)2<\/sup><\/span><\/span><\/p>\n

p2<\/sup> = 9 m2 <\/sup>+ 16 m2<\/sup><\/span><\/span><\/p>\n

p2<\/sup> = 25 m2 <\/sup><\/span><\/span><\/p>\n

p = 5 m<\/span><\/span><\/p>\n

p is a symbol of linear momentum<\/a>.<\/span><\/span><\/p>\n

5. A car is accelerated homogeneous from stationary to 22 m\/s for 9 seconds. If the tire diameter is 58 cm, specify:<\/span><\/span><\/p>\n

a) the number of rounds of tires during the movement (assume no slip)<\/span><\/span><\/p>\n

Calculate car acceleration:<\/span><\/span><\/p>\n

vt <\/sub>= vo<\/sub> + a t <\/span><\/span><\/p>\n

22 = 0 + a 9 <\/span><\/span><\/p>\n

22 = 9a <\/span><\/span><\/p>\n

a = 22\/9 = 2.44<\/span><\/span><\/p>\n

Calculate distance:<\/span><\/span><\/p>\n

d = vo<\/sub> t + \u00bd a t2<\/sup> = 0 + \u00bd a t2<\/sup><\/span><\/span><\/p>\n

d = \u00bd a t2 <\/sup>= \u00bd (2.44)(92<\/sup>) = (1.22)(81) = 98.82 m<\/span><\/span><\/p>\n

Distance of car = 98.82 meters = 9882 cm<\/span><\/span><\/p>\n

Diameter of tire = 58 cm<\/span><\/span><\/p>\n

The number of revolution = 9882 cm \/ 58 cm = 170.38 rev<\/span><\/span><\/p>\n

b) The tire’s final rotation speed in rev\/minute<\/span><\/span><\/p>\n

Final speed = 22 meters\/second = 2200 cm\/second<\/a><\/span><\/span><\/p>\n

1 rev = 1 diameter of tire = 58 cm<\/span><\/span><\/p>\n

Number of revolution:<\/span><\/span><\/p>\n

2200 cm \/ 58 cm = 40 revolutions<\/span><\/span><\/p>\n

So, the final speed in revolution\/minute<\/span><\/span><\/p>\n

40 rev\/second = 40 rev \/ 60 seconds = 0.67 rev\/minute or 0.67 rev\/minute<\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

5 Parabolic motion Work and kinetic energy Linear momentum Linear and angular motion Problems and Solutions 1. A ball is thrown from the top of a building with an initial speed of 8 m\/s at an angle of 20o below the horizontal. The ball hits the ground 3 seconds later. a) How far from 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":"5 Parabolic motion Work and kinetic energy Linear momentum Linear and angular motion Problems and Solutions","_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":"Parabolic motion Work and kinetic energy Linear momentum Linear and angular 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-4911","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\/4911","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=4911"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/4911\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=4911"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=4911"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=4911"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}