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

6.26 A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m^–1 as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest.

Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

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Payal Gupta | Contributor-Level 10

4 months ago

6.26 Mass of the block = 1 kg

Spring constant = 100 N/m

Displacement of the block, x = 10 cm = 0.1 m

At equilibrium, normal reaction, R = $m g \cos ? 37 °$

Frictional force, F = $? R$ = $m g \sin ? 37 °$

Net force acting on the block down on the incline = $m g \sin ? 37 °$ - F

= $m g \sin ? 37 °$ - $? m g \cos ? 37 °$

=mg ( $\sin ? 37 ° - ? \cos ? 37 °$ )

At equilibrium,

Work done = Potential energy of the stretched string

mg ( $\sin ? 37 ° - ? \cos ? 37 °$ ) = (1/2)kx2

1 x 10 x ( $\sin ? 37 ° - ? \cos ? 37 °$ ) = (1/2) x 100 x 0.1

10 x (0.602 – $? 0.798)$ = 0.5 x 100 x 0.1

$? = 0.128$

6.26 Mass of the block = 1 kgSpring constant = 100 N/mDisplacement of the block, x = 10 cm = 0.1 mAt equilibrium, normal reaction, R = <math><mi>m</mi><mi>g</mi><mrow><mrow><mi mathvariant="normal">cos</mi></mrow><mo>? </mo><mrow><mn>37</mn><mo>°</mo></mrow></mrow></math>Frictional force, F = <math><mi>? </mi><mi>R</mi></math> = <math><mi>m</mi><mi>g</mi><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>? </mo><mrow><mn>37</mn><mo>°</mo></mrow></mrow></math>Net force acting on the block down on the incline = <math><mi>m</mi><mi>g</mi><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>? </mo><mrow><mn>37</mn><mo>°</mo></mrow></mrow><mi></mi></math> - F= <math><mi>m</mi><mi>g</mi><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>? </mo><mrow><mn>37</mn><mo>°</mo></mrow></mrow></math> - <math><mi>? </mi><mi>m</mi><mi>g</mi><mrow><mrow><mi mathvariant="normal">cos</mi></mrow><mo>? </mo><mrow><mn>37</mn><mo>°</mo></mrow></mrow></math>=mg ( <math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>? </mo><mrow><mn>37</mn><mo>°</mo></mrow></mrow><mo>-</mo><mi></mi><mi>? </mi><mrow><mrow><mi mathvariant="normal">cos</mi></mrow><mo>? </mo><mrow><mn>37</mn><mo>°</mo></mrow></mrow></math> )At equilibrium, Work done = Potential energy of the stretched stringmg ( <math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>? </mo><mrow><mn>37</mn><mo>°</mo></mrow></mrow><mo>-</mo><mi></mi><mi>? </mi><mrow><mrow><mi mathvariant="normal">cos</mi></mrow><mo>? </mo><mrow><mn>37</mn><mo>°</mo></mrow></mrow></math> ) = (1/2)kx21 x 10 x ( <math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>? </mo><mrow><mn>37</mn><mo>°</mo></mrow></mrow><mo>-</mo><mi></mi><mi>? </mi><mrow><mrow><mi mathvariant="normal">cos</mi></mrow><mo>? </mo><mrow><mn>37</mn><mo>°</mo></mrow></mrow></math> ) = (1/2) x 100 x 0.110 x (0.602 – <math><mi></mi><mi>? </mi><mn>0.798</mn><mo>)</mo></math> = 0.5 x 100 x 0.1<math><mi>? </mi><mo>=</mo><mn>0.128</mn></math>

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

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

U_i+ K_i= U_f+ K_f

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

(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

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

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