{"id":2257,"date":"2018-04-29T06:16:15","date_gmt":"2018-04-28T22:16:15","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2257"},"modified":"2023-08-08T13:59:44","modified_gmt":"2023-08-08T13:59:44","slug":"conservation-of-mechanical-energy-on-curve-surface-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/conservation-of-mechanical-energy-on-curve-surface-problems-and-solutions.htm","title":{"rendered":"Conservation of mechanical energy on curve surface \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Conservation of mechanical energy on curve surface \u2013 problems and solutions<\/span><\/p>\n

1. If ball’s speed at point A is 6 m\/s, ball’s speed at point B is \u221a92 m\/s, and acceleration due to gravity<\/a> is g = 10 m\/s2<\/sup>. What is the height of point B (h)?<\/span><\/p>\n

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

Speed of ball at point A (vA<\/sub>) = 6 m\/s\"Conservation<\/span><\/p>\n

Speed of ball at point B (vB<\/sub>) = \u221a92 m\/s<\/span><\/p>\n

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

The height of A (hA<\/sub>) = 5.6 meters<\/span><\/p>\n

The height of B (hB<\/sub>) = h<\/span><\/p>\n

Wanted :<\/u> The height of point B (h)<\/span><\/p>\n

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

The initial mechanical energy <\/b>= the gravitational potential energy<\/b><\/span><\/p>\n

At point A, ball’s speed = 0, so the kinetic energy<\/a> of the ball = 0. KE = 1\/2 m v2 <\/sup>= 1\/2 m (0) = 0.<\/span><\/p>\n

Ball at the height of 5.6 meters so the ball has the gravitational potential energy. The gravitational potential energy<\/a> : PE = m g h = m (10)(5.6) = 56 m <\/span><\/p>\n

The initial mechanical energy<\/a> = the gravitational potential energy + kinetic energy = <\/i>56 m + 0 = 56 m <\/i><\/span><\/p>\n

The final mechanical energy = the gravitational potential energy + kinetic energy<\/b><\/span><\/p>\n

At point B, the height of ball is h. The gravitational potential energy : PE = m g h = m (10) h = 10 m h<\/span><\/p>\n

The kinetic energy of the ball : KE = 1\/2 m v2 <\/sup>= 1\/2 m (\u221a92)2 <\/sup>= 1\/2 m (92) = 46 m <\/span><\/p>\n

The final mechanical energy = the gravitational potential energy <\/i>+ the kinetic energy = 10 m h + 46 m = m (10 h + 46)<\/i><\/span><\/p>\n

The principle of conservation of mechanical energy :<\/b><\/span><\/p>\n

The initial mechanical energy = the final mechanical energy<\/i><\/span><\/p>\n

56 m = m (10 h + 46)<\/span><\/p>\n

56 = 10 h + 46<\/span><\/p>\n

56 \u2013 46 = 10 h <\/span><\/p>\n

10 = 10 h <\/span><\/p>\n

h = 10\/10 <\/span><\/p>\n

h = 1 meter<\/span><\/p>\n

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

If the initial velocity = 0, the acceleration due to gravity = 10 m\/s2<\/sup>, then what is the speed at the height of B.<\/span><\/p>\n

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

The initial speed (vo<\/sub>) = 0<\/span><\/p>\n

The initial height (ho<\/sub>) = 50 m \u2013 10 m = 40 m<\/span><\/p>\n

The final height (ht<\/sub>) = 0<\/span><\/p>\n

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

Wanted :<\/u> The final speed (vt<\/sub>)<\/span><\/p>\n

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

The change of the kinetic energy :<\/span><\/p>\n

\u0394KE = 1\/2 m (vt<\/sub>2<\/sup> \u2013 vo<\/sub>2<\/sup>) = 1\/2 m (vt<\/sub>2<\/sup> \u2013 0) = 1\/2 m vt<\/sub>2<\/sup><\/span><\/p>\n

The change of the potential energy :<\/span><\/p>\n

\u0394PE = m g (ht<\/sub> \u2013 ho<\/sub>) = m (10)(0-40) = m (10)(-40) = – 400 m <\/span><\/p>\n

Calculate the final speed (vt<\/sub>) using the equation of the principle of conservation of mechanical energy<\/a> :<\/span><\/p>\n

0 = \u0394KE + \u0394PE<\/span><\/p>\n

0 = 1\/2 m vt<\/sub>2<\/sup> – 400 m <\/span><\/p>\n

1\/2 m vt<\/sub>2<\/sup> = 400 m <\/span><\/p>\n

1\/2 vt<\/sub>2<\/sup> = 400<\/span><\/p>\n

vt<\/sub>2<\/sup> = 2(400)<\/span><\/p>\n

vt<\/sub> = \u221a(400)(2)<\/span><\/p>\n

vt<\/sub> = 20\u221a2 m\/s<\/span><\/p>\n

    \n
  1. What is meant by “conservation of mechanical energy”?<\/strong><\/span>\n