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  • 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. The work done by the net force during its displacement from x = 0 to x = 2 m is (a) 1.5 J

A body of mass 0.5 kg travels in a straight line with velocity v = a x^{\frac{3}{2}} where a = 5 m^{\frac{-1}{2}}s^{-1}. The work done by the net force during its displacement from x = 0 to x = 2 m is
(a) 1.5 J
(b) 50 J
(c) 10 J
(d) 100 J

Answers (1)

The answer is the option (b) 50 J

m = 0.5 kg

v = a x^{\frac{3}{2}}

a = 5 m^{\frac{-1}{2}}s^{-1}

now, acceleration= a = \frac{dv}{dt }= v \frac{dv}{dx}

= ax^{\frac{3}{2}} \frac{d}{dx} ax^{\frac{3}{2}} = ax^{\frac{3}{2}} \times \frac{3}{2} \times ax^{\frac{1}{2}} = \frac{3}{2} a^{2}x^{2}

Net force = ma = m \left (\frac{3}{2} a^{2}x^{2} \right )

Work done under the variable force

= \int_{x=0}^{x=2}F.dx = \int_{0}^{2}\frac{3}{2} ma^{2}x^{2} dx

= \frac{1}{2}ma^{2} \times 8 = \frac{1}{2} \times (0.5) \times (25) \times 8 = 50 J

Posted by

infoexpert24

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