{"id":2341,"date":"2018-05-01T11:24:01","date_gmt":"2018-05-01T03:24:01","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2341"},"modified":"2023-08-08T12:27:30","modified_gmt":"2023-08-08T12:27:30","slug":"work-energy-principle-nonconservative-force-motion-on-inclined-plane-with-friction-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/work-energy-principle-nonconservative-force-motion-on-inclined-plane-with-friction-problems-and-solutions.htm","title":{"rendered":"Work-energy principle Nonconservative force Motion on inclined plane with friction – Problems and Solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

2 Work-energy principle Nonconservative force Motion on inclined plane with friction – Problems and Solutions<\/strong><\/span><\/p>\n

1. A block slides down an inclined plane<\/a> with friction. What is the block’s velocity<\/a> when the block hits the ground? The coefficient of kinetic friction<\/a> is 0.4. Acceleration due to gravity is 10 m\/s2<\/sup>.<\/span><\/p>\n

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

Initial height (ho<\/sub>) = 6 m\"Work-energy<\/span><\/p>\n

Final height (ht<\/sub>) = 0 m<\/span><\/p>\n

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

Coefficient of kinetic friction (\u03bck<\/sub>) = 0.4<\/span><\/p>\n

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

cos \u03b8 = adj\/hyp = 8\/10 <\/span><\/p>\n

The vertical component of weight<\/a> = wy<\/sub> = w cos \u03b8 = m g cos \u03b8 = m (10)(8\/10) = m (10)(4\/5) = m (40\/5) = 8 m <\/span><\/p>\n

The normal force<\/a> = N = wy<\/sub> = 8 m<\/span><\/p>\n

The force of kinetic friction = fk <\/sub>= \u03bck<\/sub> N = \u03bck<\/sub> wy <\/sub>= (0.4)(8 m) = 3.2 m <\/span><\/p>\n

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

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

The work-mechanical energy principle states that the work done by the nonconservative forces acting on an object is equal to the total change in kinetic and potential energies. <\/span><\/p>\n

Wnc<\/sub> = \u0394EM\"Work-energy<\/span><\/p>\n

Wnc<\/sub> = \u0394EK + \u0394EP<\/span><\/p>\n

W<\/i>nc<\/i><\/sub> = Work done by nonconservative force, <\/i>\u0394<\/i>EM = The change in mechanical energy<\/a>, <\/i>\u0394<\/i>EK = The change in kinetic energy, <\/i>\u0394<\/i>EP = The change in potential energy.<\/i><\/span><\/p>\n

The change in kinetic energy<\/a> :<\/span><\/p>\n

\u0394EK = 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 in potential energy<\/a> :<\/span><\/p>\n

\u0394EP = m g (ht<\/sub> \u2013 ho<\/sub>) = m (10)(0-6) = m (10)(-6) = – 60 m <\/span><\/p>\n

Work done by the force<\/a> of kinetic friction :<\/span><\/p>\n

Wnc<\/sub> = – fk<\/sub> s = – (3.2 m)(10) = – 32 m<\/span><\/p>\n

Minus sign indicates that the work done by the force of kinetic friction on the block is negative.<\/i><\/span><\/p>\n

Determine the final speed (vt<\/sub>) :<\/span><\/p>\n

Wnc<\/sub> = \u0394EK + \u0394EP<\/span><\/p>\n

– 32 m = 1\/2 m vt<\/sub>2 <\/sup> – 60 m<\/span><\/p>\n

– 32 m = m (1\/2 vt<\/sub>2 <\/sup> – 60)<\/span><\/p>\n

– 32 = 1\/2 vt<\/sub>2 <\/sup> – 60<\/span><\/p>\n

– 32 + 60 = 1\/2 vt<\/sub>2<\/sup><\/span><\/p>\n

28 = 1\/2 vt<\/sub>2<\/sup><\/span><\/p>\n

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

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

vt<\/sub> = \u221a4.14<\/span><\/p>\n

vt<\/sub> = 2\u221a14 m.s-1<\/sup><\/span><\/p>\n

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

A block slides down on an inclined plane with friction. The final speed of the block when it hits the ground is 10 m\/s2<\/sup>. If the friction force is 2 N and acceleration due to gravity is 10 m\/s2<\/sup>. What is the height of h?<\/span><\/p>\n

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

Mass of block (m) = 1 kg\"Work-energy<\/span><\/p>\n

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

Final speed (vt<\/sub>) = 10 ms-1<\/sup><\/span><\/p>\n

Initial height (ho<\/sub>) = h<\/span><\/p>\n

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

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

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

Wanted :<\/u> Height (h)<\/span><\/p>\n

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

The work is done by the force of kinetic friction :<\/span><\/p>\n

Wnc<\/sub> = – fk<\/sub> d = – (2)(15) = – 30<\/span><\/p>\n

Minus sign indicates that the work done by the force of kinetic friction on the block is negative.<\/i><\/span><\/p>\n

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

\u0394EK = 1\/2 m (vt<\/sub>2<\/sup> \u2013 vo<\/sub>2<\/sup>) = 1\/2 (1)(102<\/sup> \u2013 0) = 1\/2 (102<\/sup>) = 1\/2 (100) = 50<\/span><\/p>\n

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

\u0394EP = m g (ht<\/sub> \u2013 ho<\/sub>) = (1)(10)(0-h) = (10)(-h) = -10 h<\/span><\/p>\n

The equation of the work-mechanical energy principle :<\/span><\/p>\n

Wnc<\/sub> = \u0394EK + \u0394EP<\/span><\/p>\n

– 30 = 50 – 10 h<\/span><\/p>\n

10 h = 50 + 30<\/span><\/p>\n

10 h = 80<\/span><\/p>\n

h = 80\/10<\/span><\/p>\n

h = 8 meters<\/span><\/p>\n

    \n
  1. What is the work-energy principle?<\/strong><\/span>\n