{"id":1022,"date":"2018-02-23T04:48:29","date_gmt":"2018-02-22T20:48:29","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1022"},"modified":"2018-02-23T04:48:29","modified_gmt":"2018-02-22T20:48:29","slug":"work-mechanical-energy-principle-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/work-mechanical-energy-principle-problems-and-solutions.htm","title":{"rendered":"Work-mechanical energy principle \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. A 2-kg box decelerated from 10 m\/s to rest. The coefficient of kinetic friction<\/a> is 0.2. Acceleration due to gravity<\/a> is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>. What is the magnitude of displacement<\/a>?<\/span><\/span><\/p>\n

\"Work-mechanical<\/p>\n

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

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

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 10 m\/s<\/span><\/span><\/p>\n

Final velocity (v<\/span><\/span>t<\/span><\/sub>) = 0 m\/s<\/span><\/span><\/p>\n

The coefficient of kinetic friction (<\/span><\/span>\u03bc<\/span><\/span>k<\/span><\/sub>) = 0.2<\/span><\/span><\/p>\n

Weight<\/a> (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> The magnitude of displacement (d)<\/span><\/span><\/p>\n

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

The work-mechanical energy principle<\/a> :<\/span><\/span><\/p>\n

W<\/span><\/span>nc<\/span><\/sub> = <\/span><\/span>\u0394EM<\/span><\/span><\/p>\n

W<\/b><\/span><\/span>nc<\/b><\/span><\/sub> = \u0394EK + \u0394EP<\/b><\/span><\/span><\/p>\n

W<\/b><\/span><\/span>nc<\/b><\/span><\/sub> = \u00bd m (v<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup> \u2013 v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>) + m g h<\/b><\/span><\/span><\/p>\n

W<\/i><\/span><\/span>nc<\/i><\/span><\/sub> = work<\/a> done by nonconservative force<\/a> acting on the object<\/i><\/span><\/span><\/p>\n

\u0394EK = the change in kinetic energy<\/a><\/i><\/span><\/span><\/p>\n

\u0394EP = the change in potential energy<\/a><\/i><\/span><\/span><\/p>\n

m = mass<\/i><\/span><\/span><\/p>\n

v = velocity<\/i><\/span><\/span><\/p>\n

g = acceleration due to gravity<\/i><\/span><\/span><\/p>\n

h = the change in height <\/i><\/span><\/span><\/p>\n

The change in height h = 0 so \u0394EP = 0<\/span><\/span><\/p>\n

W<\/b><\/span><\/span>nc<\/b><\/span><\/sub> = \u0394EK<\/b><\/span><\/span><\/p>\n

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

W<\/span><\/span>nc<\/span><\/sub> = -F<\/span><\/span>c<\/span><\/sub> d = – <\/span><\/span>\u03bc<\/span><\/span>k <\/span><\/sub> N d = – \u03bc<\/span><\/span>k <\/span><\/sub> w d = – \u03bc<\/span><\/span>k <\/span><\/sub> m g d <\/span><\/span><\/p>\n

W<\/span><\/span>nc<\/span><\/sub> = -(0.2)(2)(10)(s)<\/span><\/span><\/p>\n

W<\/span><\/span>nc<\/span><\/sub> = – (4)(2)<\/span><\/span><\/p>\n

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

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

\u0394EK = \u00bd m (v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> \u2013 v<\/span><\/span>o<\/span><\/sub>2<\/span><\/span><\/sup>) = \u00bd (2)(0 \u2013 10<\/span><\/span>2<\/span><\/span><\/sup>) = 0 \u2013 100 = -100<\/span><\/span><\/p>\n

The magnitude of displacement :<\/b><\/u><\/span><\/span><\/p>\n

W<\/span><\/span>nc<\/span><\/sub> = \u0394EK<\/span><\/span><\/p>\n

– 4 s = – 100<\/span><\/span><\/p>\n

s = -100\/-4<\/span><\/span><\/p>\n

s = 25 m<\/span><\/span><\/p>\n\n

2. A block slides down on inclined plane. The coefficient of kinetic friction is 0.4. Acceleration due to gravity is <\/span><\/span>g = 10 m.s<\/span><\/span>-2<\/span><\/span><\/sup>. Determine the final velocity when the block hits the ground.<\/span><\/span><\/p>\n

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

Initial height (h<\/span><\/span>o<\/span><\/sub>) = 6 m<\/span><\/span><\/p>\n

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

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 0<\/span><\/span><\/p>\n

The coefficient of kinetic friction (<\/span><\/span>\u03bc<\/span><\/span>k<\/span><\/sub>) = 0.4 <\/span><\/span><\/p>\n

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

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

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

Normal force = N = w<\/span><\/span>y<\/span><\/sub> = 8 m<\/span><\/span><\/p>\n

Force of kinetic friction = f<\/span><\/span>k <\/span><\/sub>= <\/span><\/span>\u03bc<\/span><\/span>k<\/span><\/sub> N = <\/span><\/span>\u03bc<\/span><\/span>k<\/span><\/sub> w<\/span><\/span>y <\/span><\/sub>= (0.4)(8 m) = 3.2 m <\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Final velocity (v<\/span><\/span>t<\/span><\/sub>)<\/span><\/span><\/p>\n

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

\"Work-mechanicalThe work-mechanical energy principle :<\/span><\/span><\/p>\n

W<\/span><\/span>nc <\/span><\/sub>= <\/span><\/span>\u0394<\/span><\/span>EM<\/span><\/span><\/p>\n

W<\/span><\/span>nc<\/span><\/sub> = <\/span><\/span>\u0394<\/span><\/span>EK + <\/span><\/span>\u0394<\/span><\/span>EP<\/span><\/span><\/p>\n

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

\u0394<\/span><\/span>EK = 1\/2 m (v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> \u2013 v<\/span><\/span>o<\/span><\/sub>2<\/span><\/span><\/sup>) = 1\/2 m (v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> \u2013 <\/span><\/span>0<\/span><\/span>) = 1\/2 m v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup><\/p>\n

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

\u0394<\/span><\/span>EP = m g (h<\/span><\/span>t<\/span><\/sub> \u2013 h<\/span><\/span>o<\/span><\/sub>) = m (10)(0-6) = m (10)(-6) = – 60 m <\/span><\/span><\/p>\n

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

W<\/span><\/span>nc<\/span><\/sub> = – f<\/span><\/span>k<\/span><\/sub> s = – (<\/span><\/span>3.2 m)(10) = – 32 m<\/span><\/span><\/p>\n

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

The final velocity (v<\/span><\/span>t<\/span><\/sub>) :<\/span><\/span><\/p>\n

W<\/span><\/span>nc<\/span><\/sub> = <\/span><\/span>\u0394<\/span><\/span>EK + <\/span><\/span>\u0394<\/span><\/span>EP<\/span><\/span><\/p>\n

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

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

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

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

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

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

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

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

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

<\/p>\n

3. A block slides down on rough inclined plane. The initial velocity is 0 m\/s and the final velocity is <\/span><\/span>10 ms<\/span><\/span>-1<\/span><\/span><\/sup>. If the force of kinetic friction is 2 N and acceleration due to gravity g = 10 ms<\/span><\/span>-2<\/span><\/span><\/sup>, what is height (h) ?<\/span><\/span><\/p>\n

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

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

Initial velocity (v<\/span><\/span>o<\/span><\/sub>) = 0 <\/span><\/span>(block rest)<\/i><\/span><\/span><\/p>\n

Final velocity (v<\/span><\/span>t<\/span><\/sub>) = 10 ms<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

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

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

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

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

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

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

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

W<\/span><\/span>nc<\/span><\/sub> = – f<\/span><\/span>k<\/span><\/sub> s = – (2<\/span><\/span>)(15) = – 30<\/span><\/span><\/p>\n

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

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

\u0394<\/span><\/span>EK = 1\/2 m (v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> \u2013 v<\/span><\/span>o<\/span><\/sub>2<\/span><\/span><\/sup>) = 1\/2 (1)(<\/span><\/span>10<\/span><\/span>2<\/span><\/span><\/sup> \u2013 <\/span><\/span>0<\/span><\/span>) = 1\/2 (<\/span><\/span>10<\/span><\/span>2<\/span><\/span><\/sup>) = 1\/2 (<\/span><\/span>100<\/span><\/span>) = 50<\/span><\/span><\/p>\n

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

\u0394<\/span><\/span>EP = m g (h<\/span><\/span>t<\/span><\/sub> \u2013 h<\/span><\/span>o<\/span><\/sub>) = (1)(10)(0-h) = (10)(-h) = -10 h<\/span><\/span><\/p>\n

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

W<\/span><\/span>nc<\/span><\/sub> = <\/span><\/span>\u0394<\/span><\/span>EK + <\/span><\/span>\u0394<\/span><\/span>EP<\/span><\/span><\/p>\n

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

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

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

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

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

4<\/span><\/span><\/span>. If a block moves down a roughly inclined plane, then ….
\nA. The work done by gravity force on the block is greater than the change of potential energy of the block
\nB. The mechanical energy increases
\nC. The amount of kinetic energy and its potential energy is reduced
\nD. The work done by the friction force equal to the change in kinetic energy of the block
\nSolution<\/span><\/span><\/span><\/p>\n

A is wrong<\/u><\/span><\/span><\/p>\n

Work done by the force of gravity on the block equal to the change in the gravitational potential energy of a block.<\/span><\/span><\/p>\n

B is wrong<\/u><\/span><\/span><\/p>\n

If the in<\/span><\/span>c<\/span><\/span>lined plane is smooth then the mechanical energy of the block is constant. When at the top of the inclined plane and not yet moves, the mechanical energy of the block is equal to the gravitational potential energy. The block still at rest so its kinetic energy is zero. When moves down an inclined plane, the height of the block is reduced so the gravitational potential energy is also reduced. The gravitational potential energy decreases because it changed to the kinetic energy. Although the gravitational potential energy change into the kinetic energy but the mechanical energy is constant.<\/span><\/span><\/p>\n

If the inclined plane is rough then the mechanical energy of the block is decreased because of the negative work done by the friction force. The friction force is a non-conservative force. Work done by a non-conservative force on an object causes the mechanical energy of the object is decreased.<\/span><\/span><\/p>\n

C is correct<\/u><\/span><\/span><\/p>\n

The inclined plane is rough so there is a friction force that challenges the motion of the block, The friction force is a non-conservative force. Theorem work-mechanical energy states that work done by a non-conservative force (for example friction force) equal to the change of the mechanical energy. In this chase, the mechanical energy is decreased. <\/span><\/span><\/p>\n

The mechanical energy of block = the gravitational potential energy + the kinetic energy.<\/span><\/span><\/p>\n

D is wrong.<\/u><\/span><\/span><\/p>\n

The work done by the friction force on the block equal to the change of the mechanical energy of block, not the change of the kinetic energy of the block. True that the friction force challenges the motion of the block so it decreases the speed of block and decreases the kinetic energy of the block. But realize that the kinetic energy of the block comes from the gravitational potential energy. So it’s true stated that the work done by the friction force equal to the change in the mechanical energy (mechanical energy decreases).<\/span><\/span><\/p>\n

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

5<\/span><\/span>. An object on a rough floor is hit so it moves for 3 seconds then stop. If the known mass of the object is 10 grams, the friction force between object and floor is 2 kilodyne. Determine the work done by the friction force.<\/span><\/span><\/p>\n

A. 0.18 J<\/span><\/span><\/p>\n

B. -0.18 J<\/span><\/span><\/p>\n

C. 0.36 J<\/span><\/span><\/p>\n

D. -0.36 J<\/span><\/span><\/p>\n

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

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

Final speed (v<\/span><\/span>t<\/span><\/sub>) = 0 m\/s (object rests)<\/span><\/span><\/p>\n

Mass of object (m) = 10 grams = 10\/1000 kg = 1\/100 kg = 0.01 kg<\/span><\/span><\/p>\n

Friction force (F) = 2 kilodyne = 2 x 10<\/span><\/span>3 <\/span><\/span><\/sup>dyne <\/span><\/span><\/p>\n

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

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

Conversion of unit of force :<\/span><\/span><\/p>\n

1 Newton = 1 x 10<\/span><\/span>5 <\/span><\/span><\/sup>dyne<\/span><\/span><\/p>\n

1 dyne = 1 \/ 10<\/span><\/span>5<\/span><\/span><\/sup> Newton = 1 x 10<\/span><\/span>-5 <\/span><\/span><\/sup>Newton = 10<\/span><\/span>-5<\/span><\/span><\/sup> Newton<\/span><\/span><\/p>\n

Friction force (F) = 2 x 10<\/span><\/span>3 <\/span><\/span><\/sup>dyne = 2 x 10<\/span><\/span>3<\/span><\/span><\/sup> x 10<\/span><\/span>-5<\/span><\/span><\/sup> Newton = 2 x 10<\/span><\/span>-2<\/span><\/span><\/sup> Newton = 2\/100 Newton = 0.02 Newton <\/span><\/span><\/p>\n

Theorem work-mechanical energy<\/u><\/span><\/span> states that work done by a non-conservative force on an object equal to the change in the mechanical energy of the object.<\/span><\/span><\/p>\n

If an object moves on the inclined plane then the mechanical energy (ME) = the gravitational potential energy (PE) + the kinetic energy (KE). But if the object moves just on a horizontal plane so there is no change in height the mechanical energy = the kinetic energy. On the horizontal plane, the gravitational potential energy is zero because there is no change in height.<\/span><\/span><\/p>\n

The friction force is a non-conservative force. The friction force usually decreases the object’s speed and an object’s kinetic energy. Can conclude that work done by the friction force on an object equal to the decreases of the mechanical energy.<\/span><\/span><\/p>\n

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

Work (W) = The change of the mechanical energy (\u0394EM)<\/span><\/span><\/p>\n

F s = \u0394EP + \u0394EK <\/span><\/span><\/p>\n

F s = m g \u0394h + \u00bd m (v<\/span><\/span>t<\/span><\/span><\/sub>2<\/span><\/span><\/sup> \u2013 v<\/span><\/span>o<\/span><\/span><\/sub>2<\/span><\/span><\/sup>) <\/span><\/span><\/p>\n

F s = m g (0) + \u00bd m (0<\/span><\/span>2<\/span><\/span><\/sup> \u2013 v<\/span><\/span>o<\/span><\/span><\/sub>2<\/span><\/span><\/sup>) <\/span><\/span><\/p>\n

F s = 0 + \u00bd m (\u2013 (v<\/span><\/span>o<\/span><\/span><\/sub>2<\/span><\/span><\/sup>)) <\/span><\/span><\/p>\n

F s = – \u00bd m v<\/span><\/span>o<\/span><\/span><\/sub>2 <\/span><\/span><\/sup>———— Equation 1<\/span><\/span> <\/i><\/p>\n

Description : F = friction force, d = displacement, m = mass, v<\/i><\/span><\/span>o<\/i><\/span><\/sub> = initial speed<\/i><\/span><\/span><\/p>\n

Displacement (d) and initial speed (v<\/span><\/span>o<\/span><\/sub>) not known yet because first calculate v<\/span><\/span>o<\/span><\/sub> or d. Work done by the friction force calculated using one of the equation after known v<\/span><\/span>o<\/span><\/span><\/sub> or d.<\/span><\/span><\/p>\n

The equation of displacement (d) on nonuniform linear motion :<\/span><\/span><\/p>\n

v<\/i><\/span><\/span>t<\/i><\/span><\/sub>2<\/i><\/span><\/span><\/sup> = v<\/i><\/span><\/span>o<\/i><\/span><\/sub>2<\/i><\/span><\/span><\/sup> + 2 (-a) s <\/i><\/span><\/span><\/p>\n

The final speed (v<\/i><\/span><\/span>t<\/i><\/span><\/sub>) = 0 and acceleration (a) is signed negative because the object is decelerated (the speed of object is decreases). <\/i><\/span><\/span><\/p>\n

0 = v<\/i><\/span><\/span>o<\/i><\/span><\/sub>2<\/i><\/span><\/span><\/sup> \u2013 2 a d
\n<\/i><\/span><\/span><\/p>\n

v<\/i><\/span><\/span>o<\/i><\/span><\/sub>2<\/i><\/span><\/span><\/sup> = 2 a d
\n<\/i><\/span><\/span><\/p>\n

d = v<\/i><\/span><\/span>o<\/i><\/span><\/span><\/sub>2<\/i><\/span><\/span><\/sup> \/ 2 a ———— Equation 2 <\/i><\/span><\/span><\/p>\n

Change d on equation 2 with d in equation 1 :<\/span><\/span><\/p>\n

F d = – \u00bd m v<\/span><\/span>o<\/span><\/span><\/sub>2<\/span><\/span><\/sup><\/p>\n

F (v<\/span><\/span>o<\/span><\/span><\/sub>2<\/span><\/span><\/sup>\/2a) = – \u00bd m v<\/span><\/span>o<\/span><\/span><\/sub>2<\/span><\/span><\/sup><\/p>\n

(F\/2a) v<\/span><\/span>o<\/span><\/span><\/sub>2 <\/span><\/span><\/sup>= – \u00bd m v<\/span><\/span>o<\/span><\/span><\/sub>2<\/span><\/span><\/sup><\/p>\n

F\/2a = – \u00bd m<\/span><\/span><\/p>\n

F = (2)(a)(-1\/2)(m)<\/span><\/span><\/p>\n

F = – (a)(m)<\/span><\/span><\/p>\n

a = – (F \/ m)<\/span><\/span><\/p>\n

a = – 0.02 Newton \/ 0.01 kilogram <\/span><\/span><\/p>\n

a = – 2 Newton\/kilogram<\/span><\/span><\/p>\n

a = – 2 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Equation to calculate the initial speed (v<\/span><\/span>o<\/span><\/sub>) in nonuniform linear motion :<\/span><\/span><\/p>\n

v<\/span><\/span>t<\/span><\/sub> = v<\/span><\/span>o<\/span><\/sub> + a t —–> Final speed (v<\/span><\/span>t<\/span><\/sub>) = 0, Acceleration (a) = 2 m\/s<\/span><\/span>2<\/span><\/span><\/sup>, time interval (t) = 3 seconds<\/span><\/span><\/p>\n

0 = v<\/span><\/span>o<\/span><\/sub> + (-2)(3)<\/span><\/span><\/p>\n

0 = v<\/span><\/span>o<\/span><\/sub> \u2013 6<\/span><\/span><\/p>\n

v<\/span><\/span>o<\/span><\/sub> = 6 meters\/second<\/span><\/span><\/p>\n

Work (W) done by the friction force :<\/span><\/span><\/p>\n

W = – \u00bd m v<\/span><\/span>o<\/span><\/sub>2 <\/span><\/span><\/sup>= -1\/2 (0.01)(6<\/span><\/span>2<\/span><\/span><\/sup>) = -1\/2 (0.01)(36)<\/span><\/span><\/p>\n

W = -1\/2 (0.36)<\/span><\/span><\/p>\n

W = – 0.18 Joule<\/span><\/span><\/p>\n

Work done by the friction force is signed negative means that the work decreases the mechanical energy of the object.<\/span><\/span><\/p>\n

If known d the work can calculate using the equation of <\/span><\/span>W = F d.<\/i><\/span><\/span><\/p>\n

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

[wpdm_package id=’1178′]<\/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 curved 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

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    1. A 2-kg box decelerated from 10 m\/s to rest. The coefficient of kinetic friction is 0.2. Acceleration due to gravity is 10 m\/s2. What is the magnitude of displacement? Known ; Mass (m) = 2 kg Initial velocity (vo) = 10 m\/s Final velocity (vt) = 0 m\/s The coefficient of kinetic friction (\u03bck) … 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-mechanical energy principle \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-1022","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\/1022","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=1022"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1022\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1022"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1022"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1022"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}