{"id":1011,"date":"2018-02-22T16:16:18","date_gmt":"2018-02-22T08:16:18","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1011"},"modified":"2018-02-22T16:16:18","modified_gmt":"2018-02-22T08:16:18","slug":"work-done-by-force-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/work-done-by-force-problems-and-solutions.htm","title":{"rendered":"Work done by force \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. A person pulls a block 2 m along a horizontal surface by a constant force F = 20 N. Determine the work done by force<\/a> F acting on the block. <\/span><\/span><\/p>\n

\"Work<\/p>\n

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

Force (F) = 20 N<\/span><\/span><\/p>\n

Displacement<\/a> (s) = 2 m<\/span><\/span><\/p>\n

Angle (\u03b8<\/span><\/span>) = <\/span><\/span>0<\/span><\/span><\/p>\n

Wanted<\/u><\/span><\/span> : Work (W)<\/span><\/span><\/p>\n

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

W = F d cos <\/span><\/span>\u03b8 <\/span><\/span>= (20)(2)(cos 0) = (20)(2)(1) = 40 Joule<\/span><\/span><\/p>\n

<\/p>\n

2. A force F = 10 N acting on a box 1 m along a horizontal surface. The force acts at a 30<\/span><\/span>o<\/span><\/span><\/sup> angle as shown in figure below. Determine the work done by force F!<\/span><\/span><\/p>\n

\"Work<\/p>\n

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

Force (F) = 10 N<\/span><\/span><\/p>\n

The horizontal force (F<\/span><\/span>x<\/span><\/sub>) = F cos 30<\/span><\/span>o<\/span><\/span><\/sup> = (10)(0.5\u221a3) = 5\u221a3 <\/span><\/span>N<\/span><\/span><\/p>\n

Displacement (d) = 1 meter<\/span><\/span><\/p>\n

Wanted<\/u><\/span><\/span> : Work (W) ?<\/span><\/span><\/p>\n

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

W = F<\/span><\/span>x<\/span><\/sub> d = (5\u221a3)(1) = 5\u221a3 <\/span><\/span>Joule<\/span><\/span><\/p>\n

3. A body falls freely from rest, from a height of 2 m. If acceleration due to gravity<\/a> is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>, determine the work done by the force of gravity<\/a>!<\/span><\/span><\/p>\n

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

Object’s mass<\/a> (m) = 1 kg<\/span><\/span><\/p>\n

Height (h) = 2 m<\/span><\/span><\/p>\n

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

Wanted :<\/u><\/span><\/span> Work done by the force of gravity (W)<\/span><\/span><\/p>\n

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

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

W = (1)(10)(2) = 20 Joule<\/span><\/span><\/p>\n

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

<\/p>\n

4. An 1-kg object attached to a spring so it is elongated 2 cm. If acceleration due to gravity is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>, determine (a) the spring constant (b) work done by spring force on object<\/span><\/span><\/p>\n

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

Mass (m) = 1 kg<\/span><\/span><\/p>\n

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

Elongation (x) = 2 cm = 0.02 m<\/span><\/span><\/p>\n

Weight (w) = m g = (1 kg)(10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 10 kg m\/s<\/span><\/span>2 <\/span><\/span><\/sup>= 10 N<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Spring constant and work done by spring force<\/span><\/span><\/p>\n

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

(a) Spring constant<\/u><\/span><\/span><\/p>\n

Formula of Hooke’s law<\/a> <\/b><\/span><\/span>: <\/span><\/span><\/p>\n

F = k x. <\/span><\/span><\/p>\n

k = F \/ x = w \/ x = m g \/ x<\/span><\/span><\/p>\n

k = (1)(10) \/ 0.02 = 10 \/ 0.02 <\/span><\/span><\/p>\n

k = 500 N\/m <\/span><\/span><\/p>\n

(b) work done by spring force<\/u><\/span><\/span><\/p>\n

W = – \u00bd k x<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

W = – \u00bd (500)(0.02)<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

W = – (250)(0.0004)<\/span><\/span><\/p>\n

W = -0.1 Joule<\/span><\/span><\/p>\n

The minus sign indicates that the direction of spring force is opposite with the direction of object displacement. <\/span><\/span><\/p>\n

<\/p>\n

5. A force F = 10 N accelerates a box over a displacement 2 m. The floor is rough and exerts a friction force<\/a> F<\/span><\/span>k <\/span><\/sub>= 2 N. Determine the net work done on the box. <\/span><\/span><\/p>\n

\"Work<\/p>\n

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

Force (F) = 10 N<\/span><\/span><\/p>\n

Force of kinetic friction (F<\/span><\/span>k<\/span><\/sub>) = 2 N<\/span><\/span><\/p>\n

Displacement (d) = 2 m<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Net work (W<\/span><\/span>net<\/span><\/sub>) <\/span><\/span><\/p>\n

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

Work done by force F :<\/span><\/span><\/p>\n

W<\/span><\/span>1<\/span><\/sub> = F d cos 0 = (10)(2)(1) = 20 Joule<\/span><\/span><\/p>\n

Work done by force of kinetic friction (F<\/span><\/span>k<\/span><\/sub>) :<\/span><\/span><\/p>\n

W<\/span><\/span>2<\/span><\/sub> = F<\/span><\/span>k<\/span><\/sub> d = (2)(2)(cos 180) = (2)(2)(-1) = -4 Joule<\/span><\/span><\/p>\n

Net work :<\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = W<\/span><\/span>1<\/span><\/sub> \u2013 W<\/span><\/span>2<\/span><\/sub><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = 20 \u2013 4 <\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = 16 Joule<\/span><\/span><\/p>\n

6<\/span><\/span>. What is the work done by force F on the block.<\/span><\/span><\/p>\n

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

Force (F) = 12 Newton<\/span><\/span><\/p>\n

Displacement (d) = 4 meters<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> Work (W)<\/span><\/span><\/p>\n

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

W = F d = (12 Newton)(4 meters) = 48 N m = 48 Joule<\/span><\/span><\/p>\n

<\/p>\n

7<\/span><\/span>. A block is pushed by a force of 200 N. The block’s displacement is 2 meters. What is the work done on the block?<\/span><\/span><\/p>\n

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

Force (F) = 200 Newton<\/span><\/span><\/p>\n

Displacement (d) = 2 meters<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> Work (W)<\/span><\/span><\/p>\n

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

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

W = F s <\/span><\/span><\/p>\n

W = (200 Newton)(2 meters) <\/span><\/span><\/p>\n

W = 400 N m<\/span><\/span><\/p>\n

W = 400 Joule<\/span><\/span><\/p>\n

8<\/span><\/span>. <\/span><\/span>The driver of the sedan wants to park his car exactly 0.5 m in front of the truck which is at 10 m from the sedan’s position. What is the work required by the sedan?<\/span><\/span><\/span><\/p>\n

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

Displacement (d) = 10 meters \u2013 0.5 meters = 9.5 meters<\/span><\/span><\/p>\n

Force (F) = 50 Newton<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Work (W)<\/span><\/span><\/p>\n

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

W = F s <\/span><\/span><\/p>\n

W = (50 Newton)(9.5 meters) <\/span><\/span><\/p>\n

W = 475 N m<\/span><\/span><\/p>\n

W = 475 Joule<\/span><\/span><\/p>\n

<\/p>\n

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

\"Work<\/p>\n

Work done by Tom and Jerry so the car can move as far as 4 meters. Forces exerted by Tom and Jerry are 50 N and 70 N. <\/span><\/span><\/p>\n

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

Displacement (s) = 4 meters<\/span><\/span><\/p>\n

Net force (F) = 50 Newton + 70 Newton = 120 Newton<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> Work (W)<\/span><\/span><\/p>\n

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

W = F s = (120 Newton)(4 meters) = 480 N m = 480 Joule<\/span><\/span><\/p>\n

10<\/span><\/span>. A driver pulling a car so the car moves as far as 1000 cm. What is the work done on the car?<\/span><\/span><\/p>\n

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

Force (F) = 250 Newton<\/span><\/span><\/p>\n

Displacement (s) = 1000 cm = 1000\/100 meters = 10 meters<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Work (W)<\/span><\/span><\/p>\n

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

W = F s = (250 Newton)(10 meters) = 2500 N m = 2500 Joule<\/span><\/span><\/p>\n

11. Based on figure below, if work done by net force is <\/span><\/span>375 Joule, <\/span><\/span>determine object’s displacement.<\/span><\/span><\/p>\n

\"Work<\/p>\n

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

Work <\/span><\/span>(W) = 375 Joule<\/span><\/span><\/p>\n

Net force <\/span><\/span>(<\/span><\/span>\u03a3F) = 40 N + 10 N \u2013 25 N = 25 Newton (<\/span><\/span>rightward<\/span><\/span>)<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Displacement <\/span><\/span>(<\/span><\/span>d<\/span><\/span>)<\/span><\/span><\/p>\n

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

The equation of work :<\/span><\/span><\/p>\n

W = F s<\/span><\/span><\/p>\n

Object’s displacement :<\/span><\/span><\/p>\n

d<\/span><\/span> = W \/ F = 375 Joule \/ 25 Newton<\/span><\/span><\/p>\n

d<\/span><\/span> = 15 meter<\/span><\/span>s<\/span><\/span><\/p>\n

12. The activities below w<\/span>hich <\/span>do not do <\/span>work<\/span> is<\/span> …<\/span><\/span><\/span><\/p>\n

A. Push an object as far as 10 meters<\/span><\/span><\/p>\n

B. Push a car until a move<\/span><\/span><\/p>\n

C. Push a wall<\/span><\/span><\/p>\n

D. Pulled a box<\/span><\/span><\/p>\n

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

The equation of work :<\/span><\/span><\/p>\n

W = \u03a3F s<\/span><\/span><\/p>\n

W = <\/i>work<\/i>, F = <\/i>force<\/i>, <\/i>d<\/i> = <\/i>displacement<\/i><\/span><\/span><\/p>\n

Based on the above formula, work done by force and there is a displacement.<\/span><\/span><\/p>\n

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

1<\/span><\/span>3<\/span><\/span>. Andrew pushes an object with force of 20 N so the object moves in circular motion with a radius of 7 meters. Determine the work done by Andrew for two times circular motion<\/a>.<\/span><\/span><\/p>\n

A. 0 Joule<\/span><\/span><\/p>\n

B. 1400 Joule<\/span><\/span><\/p>\n

C. 1540 Joule<\/span><\/span><\/p>\n

D. 1760 Joule<\/span><\/span><\/p>\n

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

If the person pushes <\/span><\/span>wheelchair<\/span><\/span><\/em> for two times circular motion then the person and wheelchair return to the original position, so the displacement of the person is zero. <\/span><\/span><\/em><\/p>\n

Displacement = 0 so work = 0.<\/span><\/span><\/p>\n

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

14<\/span><\/span>. Someone push an object on the floor with force of 350 N. The floor exerts a friction force 70 N. Determine the work done by force to move the object as far as 6 meters.<\/span><\/span><\/p>\n

A. 45 J <\/span><\/span><\/p>\n

B. 72 J<\/span><\/span><\/p>\n

C. 1680 J<\/span><\/span><\/p>\n

D. 2580 J<\/span><\/span><\/p>\n

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

The force of push (F) = 350 Newton<\/span><\/span><\/p>\n

Friction force (F<\/span><\/span>fric<\/span><\/sub>) = 70 Newton<\/span><\/span><\/p>\n

Displacement of object (s) = 6 meters<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> Work (W) <\/span><\/span><\/p>\n

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

There are two forces that act on the object, the push force (F) and friction force (F<\/span><\/span>fric<\/span><\/sub>). The push force has the same direction as the displacement of the object because the push force does a positive work. In another hand, the friction force has the opposite direction with a displacement of the object so that the friction force does a negative work.<\/span><\/span><\/p>\n

Work done by push force :<\/u><\/span><\/span><\/p>\n

W = F d = (350 Newton)(6 meters) = 2100 Newton-meters = 2100 Joule<\/span><\/span><\/p>\n

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

W = – (F<\/span><\/span>fric<\/span><\/sub>)(s) = – (70 Newton)(6 meters) = – 420 Newton-meters = – 420 Joule<\/span><\/span><\/p>\n

The net work :<\/u><\/span><\/span><\/p>\n

W net = 2100 Joule \u2013 420 Joule<\/span><\/span><\/p>\n

W net = 1680 Joule<\/span><\/span><\/p>\n

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

15<\/span><\/span>. An object is pushed by a horizontal force of 14 Newton on a rough floor with the friction force of 10 Newton. Determine the net work of move the object as far as 8 meters.<\/span><\/span><\/p>\n

A. 0.5 Joule<\/span><\/span><\/p>\n

B. 3 Joule<\/span><\/span><\/p>\n

C. 32 Joule<\/span><\/span><\/p>\n

D. 192 Joule<\/span><\/span><\/p>\n

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

Push force (F) = 14 Newton<\/span><\/span><\/p>\n

Friction force (F<\/span><\/span>fric<\/span><\/sub>) = 10 Newton<\/span><\/span><\/p>\n

Displacement of object (d) = 8 meters<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> Work (W) <\/span><\/span><\/p>\n

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

There are two forces that act on an object, push force (F) and friction force (F<\/span><\/span>fric<\/span><\/sub>). <\/span><\/span><\/p>\n

The push force has the same direction as the displacement of the object so that the push force does a positive work. In another hand, the friction force has the opposite direction as the displacement of the object so that the friction force does a negative work.<\/span><\/span><\/p>\n

Work done by push force :<\/u><\/span><\/span><\/p>\n

W = F s = (14 Newton)(8 meters) = 112 Newton meters = 112 Joule<\/span><\/span><\/p>\n

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

W = – (F<\/span><\/span>fric<\/span><\/sub>)(s) = – (10 Newton)(8 meters) = – 80 Newton meters = – 80 Joule<\/span><\/span><\/p>\n

The net work :<\/u><\/span><\/span><\/p>\n

W net = 112 Joule \u2013 80 Joule<\/span><\/span><\/p>\n

W net = 32 Joule<\/span><\/span><\/p>\n

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

16<\/span><\/span>. Determine the net work based on figure below.<\/span><\/span><\/p>\n

A. 360 Joule\"Work\"<\/span><\/span><\/p>\n

B. 450 Joule<\/span><\/span><\/p>\n

C. 600 Joule<\/span><\/span><\/p>\n

D. 750 Joule<\/span><\/span><\/p>\n

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

Work = Force (F) x displacement (d)<\/u><\/span><\/span><\/p>\n

Work = Area of triangle 1 + area of rectangle + area of triangle 2<\/span><\/span><\/p>\n

Work = 1\/2(40-0)(3-0) + (40-0)(9-3) + 1\/2(40-0)(12-9)<\/span><\/span><\/p>\n

Work = 1\/2(40)(3) + (40)(6) + 1\/2(40)(3)<\/span><\/span><\/p>\n

Work = (20)(3) + 240 + (20)(3)<\/span><\/span><\/p>\n

Work = 60 + 240 + 60<\/span><\/span><\/p>\n

Work = 360 Joule<\/span><\/span><\/p>\n

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

17<\/span><\/span>. <\/span><\/span>A piece of wood with a length of 60 cm plugged vertically into the ground. Wood hit with a 10-kg hammer from a height of 40 cm above the top of the wood. If the average resistance force of the ground is 2 x 10<\/span><\/span><\/span>3<\/span><\/span><\/span><\/sup> N and the acceleration due to gravity is 10 m\/s<\/span><\/span><\/span>2<\/span><\/span><\/span><\/sup>, then the wood will enter entirely into the ground after…. hits.<\/span><\/span><\/span><\/p>\n

A. 4<\/span><\/span><\/p>\n

B. 16<\/span><\/span><\/p>\n

C. 28<\/span><\/span><\/p>\n

D. 30<\/span><\/span><\/p>\n

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

Mass of hammer (m) = 10 kg<\/span><\/span><\/p>\n

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

Weight of hammer (w) = m g = (10)(10) = 100 kg m\/s<\/i><\/span><\/span>2<\/i><\/span><\/span><\/sup><\/p>\n

Displacement of hammer before hits the wood (d) = 40 cm = 0.4 meters<\/span><\/span><\/p>\n

The resistance of wood (F) = 2 x 10<\/span><\/span>3<\/span><\/span><\/sup> N = 2000 N<\/span><\/span><\/p>\n

Length of wood (s) = 60 cm = 0.6 meters<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> T<\/span><\/span><\/span>he wood will enter entirely into the ground after…. hits.<\/span><\/span><\/span><\/p>\n

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

Work done on the hammer when hammer moves as far as 0.4 meters is :<\/span><\/span><\/p>\n

W = F d = w s = (100 N)(0.4 m) = 40 Nm = 40 Joule <\/span><\/span><\/p>\n

Work done by the resistance force of the ground :<\/span><\/span><\/p>\n

W = F d = (2000 N)(0.6 m) = 1200 Nm = 1200 Joule <\/span><\/span><\/p>\n

T<\/span><\/span><\/span>he wood will enter entirely into the ground after…. hits.<\/span><\/span><\/span><\/p>\n

1200 Joule \/ 40 Joule = 30<\/span><\/span><\/p>\n

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

[wpdm_package id=’1192′]<\/p>\n

    \n
  1. Work 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 curve 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":"

    1. A person pulls a block 2 m along a horizontal surface by a constant force F = 20 N. Determine the work done by force F acting on the block. Known : Force (F) = 20 N Displacement (s) = 2 m Angle (\u03b8) = 0 Wanted : Work (W) Solution : W = … 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":"Work done by force \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-1011","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\/1011","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=1011"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1011\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1011"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1011"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1011"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}