{"id":1019,"date":"2018-02-22T18:34:26","date_gmt":"2018-02-22T10:34:26","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1019"},"modified":"2018-02-22T18:34:26","modified_gmt":"2018-02-22T10:34:26","slug":"work-and-kinetic-energy-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/work-and-kinetic-energy-problems-and-solutions.htm","title":{"rendered":"Work and kinetic energy \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Work-Kinetic energy<\/a> :<\/strong><\/p>\n

1. <\/span><\/span>A 5000-kg car accelerated from rest to 20 m\/s. Determine the net work<\/a> done on the car.<\/span><\/span><\/p>\n

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

Mass<\/a> (m) = 5000 kg<\/span><\/span><\/p>\n

Initial speed <\/span><\/span>(v<\/span><\/span>o<\/span><\/sub>) = 0 m\/s (<\/span><\/span>car rest<\/span><\/span>)<\/span><\/span><\/p>\n

Final speed <\/span><\/span>(v<\/span><\/span>t<\/span><\/sub>) = 20 m\/s<\/span><\/span><\/p>\n

Wanted<\/u><\/span><\/span> : <\/span><\/span>net work<\/span><\/span><\/p>\n

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

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

W<\/span><\/span>net<\/span><\/sub> = <\/span><\/span>\u0394<\/span><\/span>EK<\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = \u00bd m (v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> \u2013 v<\/span><\/span>o<\/span><\/sub>2<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = <\/span><\/span>net work<\/span><\/span><\/p>\n

\u0394<\/span><\/span>EK = <\/span><\/span>the change in kinetik energy<\/span><\/span><\/p>\n

m = mass (kg), <\/span><\/span><\/p>\n

v<\/span><\/span>t<\/span><\/sub> = <\/span><\/span>final speed <\/span><\/span>(m\/s), <\/span><\/span><\/p>\n

v<\/span><\/span>o<\/span><\/sub> = <\/span><\/span>initial speed <\/span><\/span>(m\/s).<\/span><\/span><\/p>\n

Net work :<\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = \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

W<\/span><\/span>net<\/span><\/sub> = \u00bd (5000)(20<\/span><\/span>2<\/span><\/span><\/sup> \u2013 0<\/span><\/span>2<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = (2500)(400 \u2013 0) <\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = (2500)(400)<\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = 1000,000 <\/span><\/span>Joule<\/span><\/span><\/p>\n

<\/p>\n

2. <\/span><\/span>A 10-kg object accelerated from 5 m\/s to 10 m\/s. Determine the net work done on the object!<\/span><\/span><\/p>\n

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

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

Initial speed <\/span><\/span>(v<\/span><\/span>o<\/span><\/sub>) = 5 m\/s<\/span><\/span><\/p>\n

Final speed <\/span><\/span>(v<\/span><\/span>t<\/span><\/sub>) = 10 m\/s<\/span><\/span><\/p>\n

Wanted<\/u><\/span><\/span> : <\/span><\/span>net work<\/span><\/span><\/p>\n

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

Net work :<\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = <\/span><\/span>\u0394<\/span><\/span>EK<\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = \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

W<\/span><\/span>net<\/span><\/sub> = \u00bd (10)(10<\/span><\/span>2<\/span><\/span><\/sup> \u2013 5<\/span><\/span>2<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = (5)(100 \u2013 25) <\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = (5)(75)<\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = 375 <\/span><\/span>Joule<\/span><\/span><\/p>\n

3. <\/span><\/span>A 2000-kg car decelerated from 10 m\/s to 5 m\/s. What is the work done on the car ?<\/span><\/span><\/p>\n

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

Car’s m<\/span><\/span>ass (m) = 2000 kg<\/span><\/span><\/p>\n

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

Final speed <\/span><\/span>(v<\/span><\/span>t<\/span><\/sub>) = 5 m\/s<\/span><\/span><\/p>\n

Wanted:<\/u> net work<\/span><\/span><\/p>\n

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

Net work :<\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = <\/span><\/span>\u0394<\/span><\/span>EK<\/span><\/span><\/p>\n

W<\/span><\/span>net <\/span><\/sub>= \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

W<\/span><\/span>net<\/span><\/sub> = \u00bd (2000)(5<\/span><\/span>2<\/span><\/span><\/sup> \u2013 10<\/span><\/span>2<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = (1000)(25 \u2013 100) <\/span><\/span><\/p>\n

W<\/span><\/span>net <\/span><\/sub>= (1000)(-75)<\/span><\/span><\/p>\n

W<\/span><\/span>net<\/span><\/sub> = -75,000 <\/span><\/span>Joule<\/span><\/span><\/p>\n

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

4. A 60-N constant force exerted on a 10-kg object for 12 seconds. The initial velocity of an object is 6 m\/s and the direction of the object is the same as the direction of the force.<\/p>\n

(1) Work done on the object is 30,240 Joule<\/p>\n

(2) The final kinetic energy is 30,240 joule<\/p>\n

(3) Power<\/a> is 2,520 Watt<\/p>\n

(4) Th increase in the kinetic energy of the object is 180 Joule<\/p>\n

The correct statements are…<\/p>\n

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

Force (F) = 60 N<\/p>\n

Time interval (t) = 12 seconds<\/p>\n

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

Initial velocity (vo<\/sub>) = 6 m\/s<\/p>\n

Wanted :<\/u> The correct statements<\/p>\n

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

Acceleration of object :<\/p>\n

\u2211F = m a<\/p>\n

60 = 10 a<\/p>\n

a = 60 \/ 10 = 6 m\/s2<\/sup><\/p>\n

The final velocity :<\/p>\n

vt<\/sub> = vo<\/sub> + a t<\/p>\n

vt<\/sub> = 6 + (6)(12)<\/p>\n

vt<\/sub> = 6 + 72<\/p>\n

vt<\/sub> = 78 m\/s<\/p>\n

The distance<\/a> traveled in 12 seconds :<\/p>\n

s = vo<\/sub> t + 1\/2 a t2 <\/sup><\/p>\n

s = (6)(12) + 1\/2 (6)(12)2 <\/sup><\/p>\n

s = 72 + (3)(144)<\/p>\n

s = 72 + 432<\/p>\n

s = 504 meters<\/p>\n

(1) Work done by force<\/p>\n

W = F s = (60)(504) = 30,240 Joule<\/p>\n

(2) The final kinetic energy<\/p>\n

KE = 1\/2 m vt<\/sub>2<\/sup> = 1\/2 (10)(78)2 <\/sup>= (5)(6084) = 30,420 Joule<\/p>\n

(3) Power<\/p>\n

P = W \/ t = 30,240 \/ 12 = 2,520 Joule\/second<\/p>\n

(4) The increase in the kinetic energy<\/p>\n

\u0394KE = 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 (10)(782 <\/sup>\u2013 62<\/sup>) = 5 (6084 \u201336) = 5 (6048)<\/p>\n

\u0394KE = 30,240 Joule<\/p>\n

<\/p>\n

5. The larger work is done by object number…<\/p>\n

\"Kinetic<\/p>\n

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

Net work = change of th kinetic energy<\/p>\n

Wnet<\/sub> = \u00bd m (vt<\/sub>2<\/sup> \u2013 vo<\/sub>2<\/sup>)<\/p>\n

The larger work :<\/p>\n

W1<\/sub> = \u00bd (8)(42<\/sup> \u2013 22<\/sup>) = (4)(16 \u2013 4) = (4)(12) = 48 Joule<\/p>\n

W2<\/sub> = \u00bd (8)(52<\/sup> \u201332<\/sup>) = (4)(25 \u2013 9) = (4)(16) = 64 Joule<\/p>\n

W3<\/sub> = \u00bd (10)(62<\/sup> \u2013 52<\/sup>) = (5)(36 \u2013 25) = (5)(11) = 55 Joule<\/p>\n

W4<\/sub> = \u00bd (10)(42<\/sup> \u2013 02<\/sup>) = (5)(16 \u2013 0) = (5)(16) = 80 Joule<\/p>\n

W5<\/sub> = \u00bd (20)(32<\/sup> \u2013 32<\/sup>) = (10)(9 \u2013 9) = (10)(0) = 0 Joule<\/p>\n

6. A 4000-kg car travels along straight line at 25 m\/s. The car is decelerated so that the car’s final velocity is 15 m\/s. What is the work done on the car.<\/p>\n

Known :<\/p>\n

Mass (m) = 4000 kg<\/p>\n

The initial velocity (vo<\/sub>) = 25 m\/s<\/p>\n

The final velocity (vt<\/sub>) = 15 m\/s<\/p>\n

Wanted <\/u>:<\/u> Work done on car<\/p>\n

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

Wnet<\/sub> = \u00bd m (vt<\/sub>2<\/sup> \u2013 vo<\/sub>2<\/sup>) = \u00bd(4000)(152<\/sup>-252<\/sup>) = (2000)(225-625) = (2000)(-400) = -800,000 Joule = -800 kJ<\/p>\n

<\/p>\n

7. A 0.1-kg thrown horizontally at 6 m\/s from the height of 5 meters. If the acceleration of gravity<\/a> is 10 m\/s2<\/sup>, then what is the kinetic energy of ball at the height of 2 meters.<\/p>\n

Known :\"Kinetic<\/u><\/p>\n

Mass (m) = 0.1 kg<\/p>\n

The change in height (h) = 5 m \u2013 2 m = 3 meters<\/p>\n

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

Wanted :<\/u> The kinetic energy at the height of 2 meters.<\/p>\n

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

Projectile motion can be understood by analyzing the horizontal and vertical components of the motion separately. Motion in horizontal direction analyzed as the constant velocity motion<\/u> and motion in vertical direction analyzed as free fall motion or vertical motion.<\/u><\/p>\n

The initial mechanical energy <\/a><\/u>= the gravitational potential energy<\/a>.<\/u><\/p>\n

PE = m g h = (0.1)(10)(3) = 3 Joule.<\/p>\n

The <\/u>final <\/u>mechanical energy <\/u>= the kinetic energy.<\/u><\/p>\n

KE = 3 Joule.<\/p>\n

8. A 1000-kg car accelerated from rest and travels at 5 m\/s. What is the work done by car?<\/p>\n

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

Mass (m) = 1000 kg<\/p>\n

Initial velocity (vo<\/sub>) = 0<\/p>\n

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

Wanted :<\/u> Work (W) done by car<\/p>\n

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

Wnet<\/sub> = \u00bd m (vt<\/sub>2<\/sup> \u2013 vo<\/sub>2<\/sup>)<\/p>\n

Work done by car :<\/p>\n

Wnet<\/sub> = \u00bd (1000)(52<\/sup> \u2013 02<\/sup>) = (500)(25 \u2013 0) = (500)(25) = 12,500 Joule<\/p>\n

<\/p>\n

9. A 500-gram ball thrown vertical upward from the surface of earth with the initial velocity 10 m\/s2<\/sup>. Acceleration due to gravity is 10 ms-2<\/sup>. What is th work done by the weight force when ball reaches the maximum height.<\/p>\n

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

Mass of ball (m) = 500 gram = 0.5 kg<\/p>\n

Initial velocity (vo<\/sub>) = 10 m\/s2<\/sup><\/p>\n

Final velocity (vt<\/sub>) = 0 (velocity at the highest point)<\/p>\n

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

Wanted :<\/u> Work (W) don by weight<\/p>\n

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

The net work done by net force on an object = the change in the kinetic energy.<\/p>\n

Wnet<\/sub> = \u0394EK = EKt <\/sub>\u2013 EKo <\/sub><\/p>\n

Wnet = \u00bd m vt<\/sub>2<\/sup> \u2013 \u00bd m vo<\/sub>2 <\/sup>= \u00bd m (vt<\/sub>2<\/sup> \u2013 vo<\/sub>2<\/sup>)<\/p>\n

KEt<\/sub> = the final kinetic energy, KEo <\/sub>= the initial kinetic energy, m = mass of object, vt<\/sub> = the final velocity of object, vo<\/sub> = initial velocity of object.<\/p>\n

Net work :<\/p>\n

Wnet <\/sub>= \u00bd m (vt<\/sub>2<\/sup> \u2013 vo<\/sub>2<\/sup>) = \u00bd (0.5)(02<\/sup> \u2013 102<\/sup>)<\/p>\n

Wnet<\/sub> = (0.25)(-100) = -25 Joule<\/p>\n

Minus sign indicates that the direction of displacement is opposite to the weight of the ball. The direction of ball is upright and the direction of weight is downright.<\/p>\n

10. A 1-kg object free fall with the height difference = 2.5 meters. Acceleration due to gravity is 10 m.s-2<\/sup>. What is the work done on the object?<\/p>\n

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

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

Initial velocity (vo<\/sub>) = 0 m\/s<\/p>\n

Height (h) = 2.5 meters<\/p>\n

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

Wanted :<\/u> Net work during displacement<\/p>\n

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

Final velocity of ball <\/u>(v<\/u>t<\/u><\/sub>)<\/u><\/p>\n

Calculated using the equation of free fall motion. Known :<\/u> Acceleration due to gravity (g) = 10 m\/s2<\/sup>, The change in height of ball (h) = 2.5 meters. Wanted :<\/u> Final velocity.<\/p>\n

vt<\/sub>2<\/sup> = 2 g h = 2(10)(2.5) = 2(25)<\/p>\n

vt<\/sub> = \u221a2(25)<\/p>\n

vt<\/sub> = 5\u221a2<\/p>\n

Net work = the change in kinetic energy<\/u><\/p>\n

Wnet<\/sub> = \u0394EK = \u00bd m (vt<\/sub>2<\/sup> \u2013 vo<\/sub>2<\/sup>) = \u00bd (1){(5\u221a2)2<\/sup> \u2013 02<\/sup>}<\/p>\n

Wnet<\/sub> = \u00bd (25)(2) = 25 Joule<\/p>\n\n

11. A 2-kg object travels at 72 km\/hour. After travels 400 meters, the final velocity of object is 144 km\/hour. Acceleration due to gravity is 10 ms-2<\/sup>. Find the net work.<\/p>\n

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

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

Initial velocity (vo<\/sub>) = 72 km\/jam = 20 m\/s<\/p>\n

Final velocity (vt<\/sub>) = 144 km\/jam = 40 m\/s<\/p>\n

Distance<\/a> (s) = 400 meters<\/p>\n

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

Wanted :<\/u> The net work<\/p>\n

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

The net work = changes of the kinetic energy<\/u><\/p>\n

Wnet <\/sub>= \u0394EK = \u00bd m (vt<\/sub>2<\/sup> \u2013 vo<\/sub>2<\/sup>) = \u00bd (2)(402<\/sup> \u2013 202<\/sup>}<\/p>\n

Wnet<\/sub> = \u00bd (2)(1600 – 400) = 1200 Joule<\/p>\n

12. A 2-kg object travels at 2 ms\u20131<\/sup>. The work done ob the object is 21 Joule. What is the final velocity of object.<\/p>\n

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

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

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

Work (W) = 21 Joule<\/p>\n

Wanted :<\/u> final velocity (vt<\/sub>)<\/p>\n

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

Wnet<\/sub>= \u0394EK<\/p>\n

Wnet<\/sub>= 1\/2 m vt<\/sub>2<\/sup> -1\/2 m vo<\/sub>2<\/sup><\/p>\n

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

21 = 1\/2 (2) (vt<\/sub>2<\/sup> \u2013 22<\/sup>)<\/p>\n

21 = (vt<\/sub>2<\/sup> \u2013 22<\/sup>)<\/p>\n

21 = vt<\/sub>2<\/sup> \u2013 4<\/p>\n

vt<\/sub>2 <\/sup>= 21 + 4 = 25<\/p>\n

vt<\/sub> = \u221a25<\/p>\n

vt<\/sub> = 5 m\/s<\/p>\n

<\/p>\n

1<\/span><\/span>3<\/span><\/span>. A 8 N constant force acts on an object with mass of 16 kg. If the object initially at rest, then determine the speed of the object after force acts on the object for 4 seconds.<\/span><\/span><\/p>\n

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

Constant force (F) = 8 Newton<\/span><\/span><\/p>\n

Mass of object (m) = 16 kg<\/span><\/span><\/p>\n

Initial speed of object (v<\/span><\/span>o<\/span><\/sub>) = 0 m\/s<\/span><\/span><\/p>\n

Time interval force acts on object (t) = 4 seconds<\/span><\/span><\/p>\n

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

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

Work = The change in the kinetic energy<\/u><\/span><\/span><\/p>\n

W = KE final \u2013 KE initial <\/span><\/span><\/p>\n

W = \u00bd m v<\/span><\/span>t<\/span><\/span><\/sub>2<\/span><\/span><\/sup> \u2013 \u00bd m v<\/span><\/span>o<\/span><\/span><\/sub>2 <\/span><\/span><\/sup><\/p>\n

W<\/span><\/span> = \u00bd m v<\/span><\/span>t<\/span><\/span><\/sub>2<\/span><\/span><\/sup> \u2013 0 <\/span><\/span><\/p>\n

W = \u00bd m v<\/span><\/span>t<\/span><\/span><\/sub>2<\/span><\/span><\/sup> —— Equation 1<\/span><\/span><\/p>\n

Work = Force x Displacement<\/u><\/span><\/span><\/p>\n

W = F d <\/span><\/span><\/p>\n

W = 8 d<\/span><\/span><\/p>\n

Use the equation of nonuniform linear motion below to calculate displacement (d) :<\/i><\/span><\/span><\/p>\n

d = v<\/i><\/span><\/span>o<\/i><\/span><\/sub> t + \u00bd a t<\/i><\/span><\/span>2<\/i><\/span><\/span><\/sup> <\/i><\/p>\n

d = displacement, v<\/i><\/span><\/span>o<\/i><\/span><\/sub> = initial velocity, t = time interval, a = acceleration<\/i><\/span><\/span><\/p>\n

d = 0 + \u00bd a t<\/i><\/span><\/span>2<\/i><\/span><\/span><\/sup> = \u00bd a t<\/i><\/span><\/span>2 <\/i><\/span><\/span><\/sup>—-> a = (v<\/i><\/span><\/span>t<\/i><\/span><\/sub> \u2013 v<\/i><\/span><\/span>o<\/i><\/span><\/sub>) \/ t = v<\/i><\/span><\/span>t<\/i><\/span><\/sub> \/ t <\/i><\/span><\/span><\/p>\n

d = \u00bd (v<\/i><\/span><\/span>t<\/i><\/span><\/sub> \/ t) t<\/i><\/span><\/span>2 <\/i><\/span><\/span><\/sup><\/p>\n

d <\/i><\/span><\/span><\/sup>= \u00bd (v<\/i><\/span><\/span>t<\/i><\/span><\/sub>) t <\/i><\/span><\/span><\/p>\n

Change displacement (d) on equation of Work with displacement (d) in this equation :<\/span><\/span><\/p>\n

W = 8 d <\/span><\/span><\/p>\n

W = 8(1\/2)(v<\/span><\/span>t<\/span><\/sub>)(t) <\/span><\/span><\/p>\n

W = (4)(v<\/span><\/span>t<\/span><\/sub>)(t) —— equation 2<\/span><\/span><\/p>\n

Equation 1 = Equation 2<\/span><\/span><\/p>\n

W = W<\/span><\/span><\/p>\n

\u00bd m v<\/span><\/span>t<\/span><\/sub>2 <\/span><\/span><\/sup>= (4)(v<\/span><\/span>t<\/span><\/sub>)(t) <\/span><\/span><\/p>\n

\u00bd m v<\/span><\/span>t<\/span> <\/sub>= (4)(t) <\/span><\/span><\/p>\n

\u00bd (16)(v<\/span><\/span>t<\/span><\/sub>) = 4(4)<\/span><\/span><\/p>\n

8 v<\/span><\/span>t<\/span><\/sub> = 16<\/span><\/span><\/p>\n

v<\/span><\/span>t<\/span><\/sub> = 16 \/ 8<\/span><\/span><\/p>\n

v<\/span><\/span>t<\/span><\/sub> = 2 meters\/second<\/span><\/span><\/p>\n

14<\/span><\/span>. To increase the speed of an object become 2 times of the initial speed, determine work required in the process…<\/span><\/span><\/p>\n

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

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

Initial speed (v<\/span><\/span>o<\/span><\/sub>) = 1 m\/s<\/span><\/span><\/p>\n

Final speed (v<\/span><\/span>t<\/span><\/sub>) = 2 x initial speed = 2 x 1 = 2 m\/s<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Work<\/span><\/span><\/p>\n

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

The initial kinetic energy :<\/span><\/span><\/p>\n

KE initial = \u00bd m v<\/span><\/span>o<\/span><\/span><\/sub>2<\/span><\/span><\/sup> = \u00bd (1)(1)<\/span><\/span>2<\/span><\/span><\/sup> = \u00bd (1)(1) = \u00bd (1) = 0.5 <\/span><\/span><\/p>\n

The final kinetic energy when the speed of object becomes 2 time of its initial speed :<\/span><\/span><\/p>\n

KE final = \u00bd m v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> = \u00bd (1)(2)<\/span><\/span>2<\/span><\/span><\/sup> = \u00bd (4) = 2 <\/span><\/span><\/p>\n

Theorem of work-kinetic energy :<\/u><\/span><\/span><\/p>\n

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

Work = The final kinetic energy\u2013 the initial kinetic energy<\/span><\/span><\/p>\n

Work = 2 \u2013 0.5<\/span><\/span><\/p>\n

Work = 1.5<\/span><\/span><\/p>\n

The initial kinetic energy = 0.5<\/span><\/span><\/p>\n

Work = 3 x 0.5 = 1.5<\/span><\/span><\/p>\n

Required work 3 times of its initial kinetic energy.<\/span><\/span><\/p>\n

15<\/span><\/span>. A car with mass of 1500 kg moves with speed of 36 km\/hour on a linear and smooth horizontal road. The car accelerated to 72 km\/hour. Determine the work required to acceleration the car.<\/span><\/span><\/p>\n

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

Mass of car (m) = 1500 kg<\/span><\/span><\/p>\n

Initial speed of car (v<\/span><\/span>o<\/span><\/sub>) = 36 km\/hour = 36,000 meters \/ 3600 second = 10 meters\/second <\/span><\/span><\/p>\n

Final speed of car (v<\/span><\/span>t<\/span><\/sub>) = 72 km\/hour = 72,000 meters \/ 3600 second = 20 meters\/second<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Work required to accelerates the car<\/span><\/span><\/p>\n

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

Theorem of work-kinetic energy :<\/span><\/span><\/p>\n

W = EK final \u2013 EK initial<\/span><\/span><\/p>\n

W = \u00bd m v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> \u2013 \u00bd m v<\/span><\/span>o<\/span><\/sub>2<\/span><\/span><\/sup> = \u00bd m (v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> \u2013v<\/span><\/span>o<\/span><\/sub>2<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

W = \u00bd (1500)(20<\/span><\/span>2<\/span><\/span><\/sup> \u2013 10<\/span><\/span>2<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

W = \u00bd (1500)(400 \u2013 100)<\/span><\/span><\/p>\n

W = \u00bd (1500)(300)<\/span><\/span><\/p>\n

W = (1500)(150)<\/span><\/span><\/p>\n

W =225,000 Joule<\/span><\/span><\/p>\n

16<\/span><\/span>. An object with mass of 2 kg initially moves at speed of 72 km.hour<\/span><\/span>-1<\/span><\/span><\/sup>. After move in horizontal straight road as far as 400 m, the speed of the object is 144 km.hour<\/span><\/span>-1<\/span><\/span><\/sup>. Determine the total work on the object.<\/span><\/span><\/p>\n

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

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

Initial speed (v<\/span><\/span>o<\/span><\/sub>) = 72 km\/hour = 72,000 meters \/ 3600 second = 20 m\/s <\/span><\/span><\/p>\n

Final speed (v<\/span><\/span>t<\/span><\/sub>) = 144 km\/hour = 144,000 meters \/ 3600 second = 40 m\/s <\/span><\/span><\/p>\n

Displacement of object = 400 meters<\/span><\/span><\/p>\n

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

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

Theorem of work-kinetic energy states that the net work acts on an object same as the change of the kinetic energy of the object.<\/span><\/span><\/p>\n

W net = KE final \u2013 KE initial<\/span><\/span><\/p>\n

W net = \u00bd m v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> \u2013 \u00bd m v<\/span><\/span>o<\/span><\/sub>2<\/span><\/span><\/sup><\/p>\n

W net = \u00bd m (v<\/span><\/span>t<\/span><\/sub>2<\/span><\/span><\/sup> \u2013 v<\/span><\/span>o<\/span><\/sub>2<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

W net = \u00bd (2)(40<\/span><\/span>2<\/span><\/span><\/sup> \u2013 20<\/span><\/span>2<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

W net = 1600 \u2013 400<\/span><\/span><\/p>\n

W net = 1200 Joule<\/span><\/span><\/p>\n

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

W = Work, KE = kinetic energy<\/span><\/span><\/p>\n

Kinetic energy<\/strong><\/p>\n

17. A 10-gram bullet moving at a constant 100 m\/s. What is the kinetic energy of the bullet.<\/span><\/span><\/p>\n

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

Mass of bullet (m) = 10 gram = 10\/1000 kilogram = 1\/100 kilogram = 0.01 kilogram<\/span><\/span><\/p>\n

Bullet’s speed (v) = 100 meters\/second<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> Kinetic energy<\/span><\/span><\/p>\n

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

KE = 1\/2 m v<\/span><\/span>2 <\/span><\/span><\/sup><\/p>\n

KE = <\/span><\/span>1\/2 (0,01 kg)(100 m\/s)<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

KE = 1\/2 (0,01 kg)(10.000 m<\/span><\/span>2<\/span><\/span><\/sup>\/s<\/span><\/span>2<\/span><\/span><\/sup>) <\/span><\/span><\/p>\n

KE = (0,01 kg)(5000 m<\/span><\/span>2<\/span><\/span><\/sup>\/s<\/span><\/span>2<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

KE = 50 kg m<\/span><\/span>2<\/span><\/span><\/sup>\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

KE = 50 Joule<\/span><\/span><\/p>\n

[wpdm_package id=’1191′]<\/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. 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 curve 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

    <\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

    Work-Kinetic energy : 1. A 5000-kg car accelerated from rest to 20 m\/s. Determine the net work done on the car. Known : Mass (m) = 5000 kg Initial speed (vo) = 0 m\/s (car rest) Final speed (vt) = 20 m\/s Wanted : net work Solution : The work-kinetic energy principle : Wnet = … 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 and kinetic energy \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-1019","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\/1019","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=1019"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1019\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1019"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1019"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1019"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}