3.0 g के कण पर एक बल इस प्रकार कार्य करता है | vedclass

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(D) दी गई स्थिति: $x = 3t - 4t^2 + t^3$वेग $v = \frac{dx}{dt} = 3 - 8t + 3t^2$त्वरण $a = \frac{dv}{dt} = -8 + 6t$द्रव्यमान $m = 3.0 \ g = 3 	imes 10^

Vedclass · Accepted Answer

$528$

Vedclass · Answer

(D) दी गई स्थिति: $x = 3t - 4t^2 + t^3$वेग $v = \frac{dx}{dt} = 3 - 8t + 3t^2$त्वरण $a = \frac{dv}{dt} = -8 + 6t$द्रव्यमान $m = 3.0 \ g = 3 	imes 10^{-3} \ kg$किया गया कार्य $W = \int F dx = \int m a v dt = \int_{0}^{4} m a v dt$$W = \int_{0}^{4} (3 	imes 10^{-3}) (-8 + 6t) (3 - 8t + 3t^2) dt$$W = 3 	imes 10^{-3} \int_{0}^{4} (-24 + 64t - 24t^2 + 18t - 48t^2 + 18t^3) dt$$W = 3 	imes 10^{-3} \int_{0}^{4} (18t^3 - 72t^2 + 82t - 24) dt$$W = 3 	imes 10^{-3} [\frac{18t^4}{4} - \frac{72t^3}{3} + \frac{82t^2}{2} - 24t]_{0}^{4}$$W = 3 	imes 10^{-3} [4.5(256) - 24(64) + 41(16) - 24(4)]$$W = 3 	imes 10^{-3} [1152 - 1536 + 656 - 96] = 3 	imes 10^{-3} [176] = 0.528 \ J = 528 \ mJ$

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