6.20 A body of mass 0.5 kg travels in a straight line with velocity v =a x 3/2 where a = 5 m–1/2 s–1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

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

**6.20 A body of mass 0.5 kg travels in a straight line with velocity v =a x ^3/2 where a = 5 m^–1/2s–1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?**

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Payal Gupta | Contributor-Level 10

5 months ago

6.20 Mass of the body = 0.5 kg

Velocity, v = a x 3/2

a = 5 m–1/2 s–1

At x =0, the initial velocity, u = 0

At x = 2, the final velocity, v = 5 $* 2^{3 / 2}$ = 14.142 m/s

Work done by the system = increase in K.E. of the body = (1/2)m ( $v^{2}$ - $u^{2}$ )

= (1/2) $* 0.5 *$ 14.142 $* 14.142$ = 50 J

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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

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

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

So v₁ = v’/ $\sqrt[]{2}$ =v²(v’/2)

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**6.20 A body of mass 0.5 kg travels in a straight line with velocity v =a x ^3/2 where a = 5 m^–1/2s–1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?**

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