{"id":2176,"date":"2018-04-26T07:22:09","date_gmt":"2018-04-25T23:22:09","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2176"},"modified":"2023-08-09T01:21:26","modified_gmt":"2023-08-09T01:21:26","slug":"potential-energy-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/potential-energy-problems-and-solutions.htm","title":{"rendered":"Potential energy \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Potential energy \u2013 problems and solutions<\/span><\/p>\n

Gravitational potential energy<\/b><\/a><\/span><\/p>\n

1. Energy 4900 Joule used to raise an object with mass of 50 kg to a height of h. What is the height of h? Acceleration due to gravity<\/a> (g) = 9.8 ms-2<\/sup>.<\/span><\/p>\n

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

Change of potential energy<\/a> (\u0394PE) = 4900 Joule<\/span><\/p>\n

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

Acceleration due to gravity (g) = 9.8 m\/s2<\/sup><\/span><\/p>\n

Wanted:<\/u> height (\u0394h)<\/span><\/p>\n

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

\u0394PE = m g \u0394h<\/span><\/p>\n

4900 = (50)(9.8) \u0394h<\/span><\/p>\n

4900 = 490 \u0394h<\/span><\/p>\n

\u0394h = 10 meters<\/span><\/p>\n

Spring potential energy<\/b><\/a><\/span><\/p>\n

2. The graph below shows the relation between force (F) and x (the change in length) of a spring. If the change in length of a spring is 8 cm, what is the spring potential energy?<\/span><\/p>\n

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

Force (F) = 2 Newton\"Potential<\/span><\/p>\n

The change in length 1 (x) = 1 cm = 1\/100 m = 0.01 m<\/span><\/p>\n

The change in length 2 = 8 cm = 8\/100 m = 0.08 m <\/span><\/p>\n

Wanted :<\/u> The spring potential energy<\/span><\/p>\n

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

Spring’s constant :<\/span><\/p>\n

\"Potential<\/span><\/p>\n

The spring potential energy :<\/span><\/p>\n

\u2206PE = 1\/2 k x2<\/sup><\/span><\/p>\n

\u2206PE = 1\/2 (200 N\/m)(0.08 m)2<\/sup><\/span><\/p>\n

\u2206PE = (100 N\/m)(0.0064 m2<\/sup>)<\/span><\/p>\n

\u2206PE = 0.64 Nm<\/span><\/p>\n

3. Based on table below, F = weight of object, \u2206L = the change in length of spring. What is the work done on the spring so the change in length of spring is 10 cm.<\/span><\/p>\n

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

The change in length of spring (\u2206L) = 10 cm = 0.1 m<\/span><\/p>\n

Wanted :<\/u> Work done on the spring<\/span><\/p>\n

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

Spring’s constant :<\/span><\/p>\n

k = F \/ \u2206x = 20 N \/ 0.04 m = 500 N\/m<\/span><\/p>\n

k = F \/ \u2206x = 30 N \/ 0.06 m = 500 N\/m <\/span><\/p>\n

k = F \/ \u2206x = 40 N \/ 0.08 m = 500 N\/m <\/span><\/p>\n

Spring’s constant is 500 N\/m<\/span><\/p>\n

The work done on the spring so the change in length of spring is 10 cm :<\/span><\/p>\n

W = 1\/2 k \u2206L2<\/sup> = 1\/2 (500 N\/m)(0.1 m)2<\/sup> = (250 N\/m)(0.01 m2<\/sup>) = 2.5 N m = 2.5 Joule.<\/span><\/p>\n

4. Graph below shows relation between force (F) and the change in length (x). What is the spring potential energy based on graph below.<\/span><\/p>\n

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

F = 40 Newton<\/span><\/p>\n

\u0394x = 0.08 m<\/span><\/p>\n

Wanted :<\/u> spring potential energy<\/span><\/p>\n

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

Spring’s constant :<\/span><\/p>\n

k = F \/ \u0394x <\/span><\/p>\n

The spring potential energy :<\/span><\/p>\n

PE = \u00bd k \u0394x2 <\/sup>= \u00bd (F\/\u0394x) \u0394x2 <\/sup>= \u00bd F \u0394x <\/span><\/p>\n

PE = \u00bd (40)(0.08) = (20)(0.08) = 1.6 Joule<\/span><\/p>\n

    \n
  1. What is potential energy?<\/strong><\/span>\n