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Work, Energy, and Power physics ncert solutions class 11th Physics Class 11th Work, Energy and Power

6.2 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.

Compute the

(a) Work done by the applied force in 10 s

(b) Work done by friction in 10 s

(c) Work done by the net force on the body in 10 s

(d) Change in kinetic energy of the body in 10 s, and interpret your results

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Payal Gupta | Contributor-Level 10

5 months ago

6.2 Given, mass of the body, m = 2 kg

Horizontal force applied, F = 7 N

Coefficient of friction, $?$ = 0.1

Acceleration, a = F/m = 7/2 = 3.5 m/s2

Frictional force, f = $? R = ? m g =$ 0.1 $* 2 * 9.807 =$ 1.96 N

Retardation produced by the frictional force, $a_{r}$ = -f/m = -1.96 /2 = 0.98 m/s2

The net acceleration by which the body moves forward

$a_{n}$ = a - $a_{r}$ = 3.5 – 0.98 = 2.52 m/s2

Distance moved by the body in 10 s is given by

s = ut + (1/2) $a_{n} t^{2}$ = 0 $* 10 + (\frac{1}{2}) * 2.52 * 10^{2}$ = 126 m

(a) Work done in 10 s is given by

W = Force $* d i s p l a c e m e n t = 7 * 126 = 882 J$

(b) Work done by friction in 10 s is given by

W = -f $* s = - 1.96 * 126 =$ - 247J

(c) Work done by the net force

...more

6.2 Given, mass of the body, m = 2 kg

Horizontal force applied, F = 7 N

Coefficient of friction, $?$ = 0.1

Acceleration, a = F/m = 7/2 = 3.5 m/s2

Frictional force, f = $? R = ? m g =$ 0.1 $* 2 * 9.807 =$ 1.96 N

Retardation produced by the frictional force, $a_{r}$ = -f/m = -1.96 /2 = 0.98 m/s2

The net acceleration by which the body moves forward

$a_{n}$ = a - $a_{r}$ = 3.5 – 0.98 = 2.52 m/s2

Distance moved by the body in 10 s is given by

s = ut + (1/2) $a_{n} t^{2}$ = 0 $* 10 + (\frac{1}{2}) * 2.52 * 10^{2}$ = 126 m

(a) Work done in 10 s is given by

W = Force $* d i s p l a c e m e n t = 7 * 126 = 882 J$

(b) Work done by friction in 10 s is given by

W = -f $* s = - 1.96 * 126 =$ - 247J

(c) Work done by the net force,

W = (F-f) $* s$ = (7 – 1.96 ) $* 126$ = 635 J

(d) Kinetic energy is given by the equation

KE = (1/2)m $v^{2}$ where v is the final velocity

From the equation v = u + at we get after 10 s, v = 0 + 2.52 $*$ 10 = 25.2 m/s

Final kinetic energy = (1/2) $* 2 * {25.2}^{2}$ = 635 J

Initial kinetic energy = (1/2) $* 2 * 0^{2}$ = 0

Change in kinetic energy = 635 – 0 = 635 J

Hence the work done by the net force = change in kinetic energy

less

6.2 Given, mass of the body, m = 2 kgHorizontal force applied, F = 7 NCoefficient of friction, <math><mi>?</mi></math> = 0.1Acceleration, a = F/m = 7/2 = 3.5 m/s2Frictional force, f = <math><mi>?</mi><mi>R</mi><mo>=</mo><mi></mi><mi>?</mi><mi>m</mi><mi>g</mi><mo>=</mo></math> 0.1 <math><mo>×</mo><mn>2</mn><mo>×</mo><mn>9.807</mn><mo>=</mo></math> 1.96 NRetardation produced by the frictional force, <math><msub><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mi>r</mi></mrow></mrow></msub></math> = -f/m = -1.96 /2 = 0.98 m/s2The net acceleration by which the body moves forward<math><msub><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mi>n</mi></mrow></mrow></msub></math> = a - <math><msub><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mi>r</mi></mrow></mrow></msub></math> = 3.5 – 0.98 = 2.52 m/s2Distance moved by the body in 10 s is given bys = ut + (1/2) <math><msub><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mi>n</mi></mrow></mrow></msub><msup><mrow><mrow><mi>t</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = 0 <math><mo>×</mo><mn>10</mn><mo>+</mo><mfenced separators="|"><mrow><mrow><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></mrow></mrow></mfenced><mo>×</mo><mn>2.52</mn><mi></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = 126 m(a) Work done in 10 s is given byW = Force <math><mo>×</mo><mi mathvariant="normal">d</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">s</mi><mi mathvariant="normal">p</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">m</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">t</mi><mo>=</mo><mn>7</mn><mi></mi><mo>×</mo><mn>126</mn><mo>=</mo><mn>882</mn><mi></mi><mi mathvariant="normal">J</mi></math> (b) Work done by friction in 10 s is given byW = -f <math><mo>×</mo><mi>s</mi><mo>=</mo><mi></mi><mo>-</mo><mn>1.96</mn><mi></mi><mo>×</mo><mn>126</mn><mo>=</mo></math> - 247J (c) Work done by the net force,W = (F-f) <math><mo>×</mo><mi>s</mi></math> = (7 – 1.96 ) <math><mo>×</mo><mn>126</mn><mi></mi></math> = 635 J (d) Kinetic energy is given by the equationKE = (1/2)m <math><msup><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> where v is the final velocityFrom the equation v = u + at we get after 10 s, v = 0 + 2.52 <math><mo>×</mo></math> 10 = 25.2 m/sFinal kinetic energy = (1/2) <math><mo>×</mo><mn>2</mn><mi></mi><mo>×</mo><msup><mrow><mrow><mn>25.2</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = 635 JInitial kinetic energy = (1/2) <math><mo>×</mo><mn>2</mn><mi></mi><mo>×</mo><msup><mrow><mrow><mn>0</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = 0Change in kinetic energy = 635 – 0 = 635 JHence the work done by the net force = change in kinetic energy

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(b) conserving energy between “O” ans ”A”

U_i+ K_i= U_f+ K_f

0+1/2mv²= mgh + 1/2mv’

$\frac{{(v')}^{2}}{2} = \frac{v^{2}}{2} = - g h$

(v’)²=v²-2gh = v^’= $\sqrt[]{v^{2} - 2 g h}$ ……….1

Let speed after emerging be v₁then

=1/2mv₁²=1/2[1/2mv’²]

1/2m(v₁)²=1/4m(v’)²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^{2}}{2} - g h}$ ………….2

From eqn 1 and 2

$\frac{v'}{v_{1}} = \frac{\sqrt[]{v^{2} - 2 g h}}{\sqrt[]{v^{2} - 2 g h / \sqrt[]{2}}} = \sqrt[]{2}$

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