{"id":2009,"date":"2018-04-21T11:06:53","date_gmt":"2018-04-21T03:06:53","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2009"},"modified":"2023-08-09T04:39:40","modified_gmt":"2023-08-09T04:39:40","slug":"dynamics-object-connected-by-cord-over-pulley-atwood-machine-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/dynamics-object-connected-by-cord-over-pulley-atwood-machine-problems-and-solutions.htm","title":{"rendered":"Dynamics Object connected by cord over pulley Atwood machine – Problems and Solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

10 Dynamics Object connected by cord over pulley Atwood machine – Problems and Solutions<\/strong><\/p>\n

1. Block A with a mass<\/a> of 5 kg, placed on a smooth horizontal plane. Block B with a mass of 3 kg hanging at one end of the cord connected with block A over a pulley. Acceleration due to gravity<\/a> is 10 m\/s2<\/sup>. What is the acceleration<\/a> of both blocks?<\/span><\/span><\/p>\n

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

Mass of block A (mA<\/sub>) = 5 kg\"Dynamics,<\/span><\/span><\/p>\n

Mass of block B (mB<\/sub>) = 3 kg<\/span><\/span><\/p>\n

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

Weight<\/a> of block B (wB<\/sub>) = mB<\/sub> g = (3)(10) = 30 Newton <\/span><\/span><\/p>\n

Wanted:<\/u> Acceleration of both blocks (a)<\/span><\/span><\/p>\n

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

The horizontal plane is smooth so there is no friction force. The force that accelerates both blocks is the weight of the block B.<\/span><\/span><\/p>\n

\u03a3F = m a<\/span><\/span><\/p>\n

wB<\/sub> = (mA<\/sub> + mB<\/sub>) a<\/span><\/span><\/p>\n

30 = (5 + 3) a<\/span><\/span><\/p>\n

30 = 8 a<\/span><\/span><\/p>\n

a = 30 \/ 8<\/span><\/span><\/p>\n

a = 3.75 m\/s2<\/sup><\/span><\/span><\/p>\n

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

Based on the figure above, \"Dynamics,<\/span><\/span><\/p>\n

(1) acceleration of object = 0<\/span><\/span><\/p>\n

(2) the object moves at a constant velocity<\/span><\/span><\/p>\n

(3) object at rest<\/span><\/span><\/p>\n

(4) the object moves if the weight of the object is smaller than the force that pulls the object.<\/span><\/span><\/p>\n

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

(1) The acceleration of the object = 0. <\/span><\/span><\/p>\n

Net force :<\/i><\/span><\/span><\/p>\n

\u2211F = m a –> <\/i>acceleration <\/i>(a) = 0<\/i><\/span><\/span><\/p>\n

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

F<\/i>1<\/i><\/sub> + F<\/i>2<\/i><\/sub> \u2013 F<\/i>3<\/i><\/sub> = 12 + 24 \u2013 36 = 36 \u2013 36 = 0 N<\/i><\/span><\/span><\/p>\n

(2) The object moves at a constant velocity<\/span><\/span><\/p>\n

No acceleration means object at rest or moves at a constant velocity. <\/i><\/span><\/span><\/p>\n

(3) object at rest<\/span><\/span><\/p>\n

No net force means object at rest.<\/i><\/span><\/span><\/p>\n

(4) the object moves if the weight of the object is smaller than the force that pulls the object.<\/span><\/span><\/p>\n

Weight acts on the vertical direction, while the pull force acts on the horizontal direction.<\/i><\/span><\/span><\/p>\n

The object moves in a horizontal direction so only the horizontal forces that act on the object<\/i><\/span><\/span><\/p>\n

3. If the coefficient of kinetic friction<\/a> between the block A and the table surface is 0.1. Acceleration due to gravity is 10 m\/s2, then what is the force acts on the block A so that the system moves to leftward in 2 m\/s2<\/sup>.<\/span><\/span><\/p>\n

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

Mass of block A (mA<\/sub>) = 30 kg\"Dynamics,<\/span><\/span><\/p>\n

weight of block <\/i>A (w<\/i>A<\/i><\/sub>) = (30 kg)(10 m\/s<\/i>2<\/i><\/sup>) = 300 kg m\/s<\/i>2<\/i><\/sup> <\/i>or <\/i>300 Newton<\/i><\/span><\/span><\/p>\n

Mass of block B (mB<\/sub>) = 20 kg<\/span><\/span><\/p>\n

weight of block <\/i>B (w<\/i>B<\/i><\/sub>) = (20 kg)(10 m\/s<\/i>2<\/i><\/sup>) = 200 kg m\/s<\/i>2 <\/i><\/sup>or <\/i>200 Newton<\/i><\/span><\/span><\/p>\n

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

Coefficient of kinetic friction (\u03bc<\/span>k<\/sub>) = 0.1<\/span><\/span><\/p>\n

Acceleration of system (a) = 2 m\/s2<\/sup> (to leftward)<\/span><\/span><\/p>\n

Force of kinetic friction <\/i>(f<\/i>k<\/i><\/sub>) = <\/i>\u03bc<\/i><\/span>k <\/i><\/sub>N = <\/i>\u03bc<\/i><\/span>k <\/i><\/sub>w<\/i>A<\/i><\/sub> = (0.1)(300) = 30 Newton <\/i><\/span><\/span><\/p>\n

Wanted :<\/u> The magnitude of force F<\/span><\/span><\/p>\n

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

Newton’s second law :<\/span><\/span><\/p>\n

\u03a3F <\/span>= m a<\/span><\/span><\/p>\n

The object A moves to leftward :<\/span><\/span><\/p>\n

F \u2013 fk<\/sub> \u2013 wB<\/sub> = (mA<\/sub> + mB<\/sub>) a<\/span><\/span><\/p>\n

F \u2013 30 \u2013 200 = (30 + 20)(2)<\/span><\/span><\/p>\n

F \u2013 230 = (50)(2)<\/span><\/span><\/p>\n

F \u2013 230 = 100<\/span><\/span><\/p>\n

F = 230 + 100<\/span><\/span><\/p>\n

F = 330 Newton<\/span><\/span><\/p>\n

4. Two objects, A = 2 kg and B = 6 kg, attached at one end of cord over a pulley, as shown in figure below. If acceleration due to gravity is 10 ms-2<\/sup> then what is the acceleration of the object B.<\/span><\/span><\/p>\n

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

Mass of object A (mA<\/sub>) = 2 kg, mB<\/sub> = 6 kg, g = 10 m\/s2\"Dynamics,<\/sup><\/span><\/span><\/p>\n

weight of object A (wA<\/sub>) = (mA<\/sub>)(g) = (2)(10) = 20 N<\/span><\/span><\/p>\n

weight of object B (wB<\/sub>) = (mB<\/sub>)(g) = (6)(10) = 60 N<\/span><\/span><\/p>\n

Wanted :<\/u> Acceleration of object b (system’s acceleration).<\/span><\/span><\/p>\n

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

wB<\/sub> > wA<\/sub> so that object B moves downward, object A moves upward <\/span><\/span><\/p>\n

\u03a3F <\/span>= m a<\/span><\/span><\/p>\n

wB<\/sub> \u2013 wA<\/sub> = (mA<\/sub> + mB<\/sub>) a<\/span><\/span><\/p>\n

60 \u2013 20 = (2 + 6) a<\/span><\/span><\/p>\n

40 = (8) a<\/span><\/span><\/p>\n

a = 5 m\/s2<\/sup><\/span><\/span><\/p>\n

5. Two objects connected by a cord over a smooth pulley, as shown in figure below. If m1<\/sub> = 1 kg, m2<\/sub> = 2 kg, and acceleration due to gravity is 10 ms-2<\/sup>, then what is the tension force T.<\/span><\/span><\/p>\n

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

Mass of object 1 (m1<\/sub>) = 1 kg\"Dynamics,<\/span><\/span><\/p>\n

Mass of object 2 (m2<\/sub>) = 2 kg<\/span><\/span><\/p>\n

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

weight of object 1 (w1<\/sub>) = m1<\/sub> g = (1 kg)(10 m\/s2<\/sup>) = 10 kg m\/s2<\/sup> or 10 Newton<\/span><\/span><\/p>\n

weight of object 2 (w2<\/sub>) = m2 <\/sub>g = (2 kg)(10 m\/s2<\/sup>) = 20 kg m\/s2<\/sup> or 20 Newton<\/span><\/span><\/p>\n

Wanted :<\/u> The tension force (T) ?<\/span><\/span><\/span><\/p>\n

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

w<\/span>2<\/span><\/sub> > w<\/span>1<\/span> <\/sub>so <\/span>m<\/span>2<\/sub><\/span> moves downward<\/span>, m<\/span>1<\/sub><\/span> moves upward<\/span>.<\/span><\/span><\/span><\/p>\n

The Newton’s second law of motion<\/a> :<\/span><\/span><\/span><\/p>\n

\u03a3F <\/span><\/span>= m a<\/span><\/span><\/span><\/p>\n

w<\/span>2<\/span><\/sub> \u2013 w<\/span>1<\/span><\/sub> = (m<\/span>1<\/span><\/sub> + m<\/span>2<\/span><\/sub>) a<\/span><\/span><\/span><\/p>\n

20 \u2013 10 = (1 + 2 ) a<\/span><\/span><\/span><\/p>\n

10 = (3) a <\/span><\/span><\/span><\/p>\n

a = 3.3 m\/s<\/span>2<\/sup><\/span><\/span><\/span><\/p>\n

System’s acceleration = <\/span>3.3 m\/s<\/span>2<\/sup><\/span>.<\/span><\/span><\/span><\/p>\n

m<\/span>2<\/sub><\/span> moves downward :<\/span><\/span><\/span><\/p>\n

w<\/span>2<\/span><\/sub> \u2013 T<\/span>2<\/span><\/sub> = m<\/span>2<\/span><\/sub> a<\/span><\/span><\/span><\/p>\n

20 \u2013 T<\/span>2<\/span><\/sub> = (2)(3.33)<\/span><\/span><\/span><\/p>\n

20 \u2013 T<\/span>2<\/span><\/sub> = 6.66<\/span><\/span><\/span><\/p>\n

T<\/span>2<\/span><\/sub> = 20 \u2013 6.66 <\/span><\/span><\/span><\/p>\n

T<\/span>2<\/span><\/sub> = 13.3 Newton<\/span><\/span><\/span><\/p>\n

m<\/span>1<\/span><\/sub> moves upward :<\/span><\/span><\/span><\/p>\n

T<\/span>1<\/span><\/sub> \u2013 w<\/span>1<\/span><\/sub> = m<\/span>1<\/sub><\/span> a<\/span><\/span><\/span><\/p>\n

T<\/span>1<\/span><\/sub> \u2013 10 = (1)(3.3)<\/span><\/span><\/span><\/p>\n

T<\/span>1<\/span><\/sub> \u2013 10 = 3.33<\/span><\/span><\/span><\/p>\n

T<\/span>1<\/span><\/sub> = 10 + 3.33<\/span><\/span><\/span><\/p>\n

T<\/span>1<\/span><\/sub> = 13.3 Newton<\/span><\/span><\/span><\/p>\n

The tension force (T) = 13.3 Newton.<\/span><\/span><\/span><\/p>\n

6. Mass of m1<\/sub> = 6 kg and mass of m2<\/sub> = 4 kg. The horizontal surface is smooth. Acceleration due to gravity is 10 m\/s2<\/sup>. What is the system’s acceleration.<\/span><\/span><\/p>\n

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

Mass of m1<\/sub> = 6 kg\"Dynamics,<\/span><\/span><\/p>\n

mass of m2 <\/sub>= 4 kg<\/span><\/span><\/p>\n

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

weight of w1 <\/sub>= m1 <\/sub>g = (6 kg)(10 m\/s2<\/sup>) = 60 kg m\/s2<\/sup> or 60 Newton<\/span><\/span><\/p>\n

weight of w2<\/sub> = m2<\/sub> g = (4 kg)(10 m\/s2<\/sup>) = 40 kg m\/s2<\/sup> or 40 Newton<\/span><\/span><\/p>\n

Wanted :<\/u> System’s acceleration (a)<\/span><\/span><\/p>\n

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

m1<\/sub> on a smooth horizontal plane without friction so that the system accelerated by the weight of the block 2.<\/span><\/span><\/p>\n

Apply Newton’s second law :<\/span><\/span><\/p>\n

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

w2<\/sub> = (m1<\/sub> + m2<\/sub>) a<\/span><\/span><\/p>\n

40 N = (6 kg + 4 kg) a<\/span><\/span><\/p>\n

40 N = (10 kg) a<\/span><\/span><\/p>\n

a = 40 N \/ 10 kg<\/span><\/span><\/p>\n

a = 4 m\/s2<\/sup><\/span><\/span><\/p>\n

7. Two blocks, each block has the mass of 2 kg, connected by a cord over a pulley, as shown in the figure below. The horizontal plane and pulley are smooth. If the block B is pulled by a horizontal force of 40 Newton, then what is the acceleration of block. Acceleration due to gravity is 10 m\/s2<\/sup>.<\/span><\/span><\/p>\n

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

mass of block A (mA<\/sub>) = mass of block B (mB<\/sub>) = 2 kg\"Dynamics,<\/span><\/span><\/p>\n

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

Force of F = 40 N<\/span><\/span><\/p>\n

weight of object A (wA<\/sub>) = m g = (2)(10) = 20 N<\/span><\/span><\/p>\n

Wanted :<\/u> System’s acceleration (a) ?<\/span><\/span><\/p>\n

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

Apply Newton’s second law :<\/span><\/span><\/p>\n

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

F – wA<\/sub> = (mA<\/sub> + mB<\/sub>) a<\/span><\/span><\/p>\n

40 – 20 = (2 + 2) a<\/span><\/span><\/p>\n

20 = (4) a<\/span><\/span><\/p>\n

a = 20 \/ 4<\/span><\/span><\/p>\n

a = 5 m\/s2<\/sup><\/span><\/span><\/p>\n

8. Mass of block A = 2 kg and mas of block B = 1 kg. Block B initially at rest, then accelerated downward until it hits ground. Acceleration due to gravity is 10 m\/s2<\/sup>. What is the magnitude of the tension force.<\/span><\/span><\/p>\n

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

Mass of block A (mA<\/sub>) = 2 kg\"Dynamics,<\/span><\/span><\/p>\n

Mass of block B (mB<\/sub>) = 1 kg<\/span><\/span><\/p>\n

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

Weight of block B (wB<\/sub>) = mB<\/sub> g = (1)(10) = 10 Newton<\/span><\/span><\/p>\n

Wanted :<\/u> The magnitude of the tension force (T)<\/span><\/span><\/p>\n

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

Ignore the friction force.<\/span><\/span><\/p>\n

System’s acceleration (a)<\/u><\/span><\/span><\/p>\n

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

wB<\/sub> = (mA <\/sub>+ mB<\/sub>) a<\/span><\/span><\/p>\n

10 = (2 + 1) a<\/span><\/span><\/p>\n

10 = 3 a<\/span><\/span><\/p>\n

a = 10\/3 <\/span><\/span><\/p>\n

The tension force <\/u>(T)<\/u><\/span><\/span><\/p>\n

The tension force on the block A :<\/span><\/span><\/p>\n

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

T = mA<\/sub> a = (2)(10\/3) = 20\/3 = 6.7 Newton<\/span><\/span><\/p>\n

The tension force on the block B :<\/u><\/span><\/span><\/p>\n

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

wB<\/sub> – T = mB<\/sub> a <\/span><\/span><\/p>\n

10 \u2013 T = (1)(10\/3)<\/span><\/span><\/p>\n

10 \u2013 T = 3.3<\/span><\/span><\/p>\n

T = 10 \u2013 3.3 = 6.7 Newton <\/span><\/span><\/p>\n

The tension force (T) = 6.7 Newton<\/span><\/span><\/p>\n

9. Mass of block A = 2 kg and mass of block B = 1 kg. The friction force between object A with the horizontal plane = 2.5 Newton. Ignore friction on pulley and cord. What is the acceleration of both blocks.<\/span><\/span><\/p>\n

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

Mass of block A (mA<\/sub>) = 2 kg\"Dynamics,<\/span><\/span><\/p>\n

Mass of block B (mB<\/sub>) = 1 kg<\/span><\/span><\/p>\n

Friction force between lock a and the horizontal plane (fkA<\/sub>) = 2.5 Newton<\/span><\/span><\/p>\n

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

Weight of block (wB<\/sub>) = mB<\/sub> g = (1)(10) = 10 Newton<\/span><\/span><\/p>\n

Wanted :<\/u> Acceleration of both blocks (a)<\/span><\/span><\/p>\n

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

Apply Newton’s second law :<\/span><\/span><\/p>\n

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

wB<\/sub> \u2013 fk<\/sub> = (mA<\/sub> + mB<\/sub>) a<\/span><\/span><\/p>\n

10 \u2013 2.5 = (2 + 1) a<\/span><\/span><\/p>\n

7.5 = 3 a<\/span><\/span><\/p>\n

a = 7.5 \/ 3 <\/span><\/span><\/p>\n

a = 2.5 m\/s2<\/sup><\/span><\/span><\/p>\n

10. Mass of block a = 30 kg, rest on a horizontal plane connected with block B with mass of 10 kg over a pulley. What is the system’s acceleration. Acceleration due to gravity is 10 ms-2<\/sup>.<\/span><\/span><\/p>\n

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

Mass of block A (mA<\/sub>) = 30 kg\"Dynamics,<\/span><\/span><\/p>\n

Mass of block B (mB<\/sub>) = 10 kg<\/span><\/span><\/p>\n

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

weight of block B (wB<\/sub>) = mB<\/sub> g = (10)(10) = 100 Newton <\/span><\/span><\/p>\n

Wanted :<\/u> System’s acceleration (a)<\/span><\/span><\/p>\n

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

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

wB<\/sub> = (mA<\/sub> + mB<\/sub>) a<\/span><\/span><\/p>\n

100 = (30 + 10) a<\/span><\/span><\/p>\n

100 = 40 a<\/span><\/span><\/p>\n

a = 100 \/ 40 <\/span><\/span><\/p>\n

a = 2.5 m\/s2<\/sup><\/span><\/span>\u00a0<\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

10 Dynamics Object connected by cord over pulley Atwood machine – Problems and Solutions 1. Block A with a mass of 5 kg, placed on a smooth horizontal plane. Block B with a mass of 3 kg hanging at one end of the cord connected with block A over a pulley. Acceleration due to gravity … 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":"Dynamics object connected by cord over pulley atwood machine - 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-2009","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\/2009","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=2009"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2009\/revisions"}],"predecessor-version":[{"id":8669,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2009\/revisions\/8669"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=2009"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=2009"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=2009"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}