{"id":2166,"date":"2018-04-26T06:01:12","date_gmt":"2018-04-25T22:01:12","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2166"},"modified":"2023-08-09T01:36:43","modified_gmt":"2023-08-09T01:36:43","slug":"net-work-gravitational-potential-energy-kinetic-energy-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/net-work-gravitational-potential-energy-kinetic-energy-problems-and-solutions.htm","title":{"rendered":"Net work Gravitational potential energy Kinetic energy \u2013 Problems and Solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

4 Net work Gravitational potential energy Kinetic energy \u2013 Problems and Solutions<\/strong><\/span><\/p>\n

1. A 5-kg object at the height of 10-meters above the ground. Acceleration due to gravity<\/a> is 10 m\/s2<\/sup>. What is the work<\/a> done on the object to moves it upward to the height of 15-meters above the ground?<\/span><\/p>\n

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

Mass<\/a> of object (m) = 5 kg<\/span><\/p>\n

Height (h) = 10 meters<\/span><\/p>\n

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

Wanted:<\/u> Work was done on the object to moves it upward to the height of 15-meters above the ground.<\/span><\/p>\n

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

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

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

Work done on the object :<\/span><\/p>\n

W = m g h = (5 kg)(10 m\/s2<\/sup>)(10 m) = 500 kg m2<\/sup>\/s2<\/sup> = 500 Joule<\/span><\/p>\n

2. A 2-kg object moves along smooth horizontal surface with the speed of 2 m\/s. If the final speed of the object is 5 m\/s. What is the work done on the object.<\/span><\/p>\n

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

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

Initial velocity (vo<\/sub>) = 2 m\/s<\/span><\/p>\n

Final velocity (vt<\/sub>) = 5 m\/s<\/span><\/p>\n

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

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

The work-kinetic energy principle<\/a> states that the net work done on an object is equal to the change in the object’s kinetic energy<\/a>. <\/span><\/p>\n

W = \u0394KE = KE2<\/sub> \u2013 KE1<\/sub> = 1\/2 m vt<\/sub>2<\/sup> \u2013 1\/2 m vo<\/sub>2 <\/sup>= 1\/2 m (vt<\/sub>2<\/sup> \u2013 vo<\/sub>2<\/sup>) = 1\/2 (2)(52<\/sup>-22<\/sup>) = (1)(25-4) = 21 Joule<\/span><\/p>\n

3. An object with mass of 10-kg initially at rest on a smooth floor. After pushed in 3 seconds, the object accelerated at 2 m\/s2<\/sup>. What is the work done on the object.<\/span><\/p>\n

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

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

Initial velocity (vo<\/sub>) = 0 (initially at rest)<\/span><\/p>\n

Time interval (t) = 3 seconds<\/span><\/p>\n

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

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

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

First, determine distance using the equation of the constant acceleration<\/a> motion.<\/span><\/p>\n

d = vo<\/sub> t + \u00bd a t2 <\/sup>= (0)(3) + \u00bd (2)(3)2<\/sup> <\/span><\/p>\n

d = 0 + (1)(9) = 9<\/span><\/p>\n

Work done on the object :<\/span><\/p>\n

W = F s = (m a)(d) = (10)(2)(9) = 180 Joule<\/span><\/p>\n

4. A block with mass of 1.5 kg accelerated upward by a constant force F = 15 N on a inclined plane as shown in figure below. Acceleration due to gravity is 10 m\/s2<\/sup> and no friction between block and inclined plane<\/a>. What is the net work done on the object.<\/span><\/p>\n

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

Mass of block (m) = 1.5 kg\"Net-work-gravitational-potential-energy-kinetic-energy-\u2013-problems-and-solutions-1\"<\/span><\/p>\n

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

Weight of block (w) = m g = (1.5)(10) = 15 Newton<\/span><\/p>\n

Force (F) = 15 Newton<\/span><\/p>\n

Wanted:<\/u> Net work is done on the object <\/span><\/p>\n

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

There are two forces do work on the block. First, a force parallel to the block’s displacement that is Force F. Second, horizontal component of the weight of block (wx<\/sub>).<\/span><\/p>\n

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

W1<\/sub> = F s = (15 N)(2 m) = 30 N m = 30 Joule<\/span><\/p>\n

Work done by horizontal component of weight (wx<\/sub>) :<\/span><\/p>\n

W2<\/sub> = wx<\/sub> s = (w sin 30o<\/sup>)(2 m) = (15 N)(1\/2)(2 m) = 15 N m = 15 Joule<\/span><\/p>\n

Net work done on the object :<\/span><\/p>\n

Wnet<\/sub> = W1<\/sub> \u2013 W2<\/sub> = 30 J \u2013 15 J = 15 Joule<\/span><\/p>\n

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  1. What is the work-energy theorem in terms of net work and kinetic energy?<\/strong><\/span>\n