A bullet of mass 1 kg fired with a speed 2 | vedclass

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(D) Given:Mass of bullet $m = 1 \ kg$Initial velocity $u = 2 \ m \ s^{-1}$Retarding force $F = -0.5/x$According to the work-energy theorem, the work d

Vedclass · Accepted Answer

$19.3$

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(D) Given:Mass of bullet $m = 1 \ kg$Initial velocity $u = 2 \ m \ s^{-1}$Retarding force $F = -0.5/x$According to the work-energy theorem, the work done by the force equals the change in kinetic energy:$W = \Delta KE = K_f - K_i$Since the bullet stops, $K_f = 0$, so $W = -K_i = -\frac{1}{2} m u^2$$W = -\frac{1}{2} 	imes 1 	imes (2)^2 = -2 \ J$Also, $W = \int_{x_1}^{x_2} F \ dx = \int_{10-L/2}^{10+L/2} -\frac{0.5}{x} \ dx$$-0.5 [\ln(x)]_{10-L/2}^{10+L/2} = -2$$\ln \left( \frac{10+L/2}{10-L/2} ight) = \frac{-2}{-0.5} = 4$$\frac{10+L/2}{10-L/2} = e^4 = 55$$10 + L/2 = 55(10 - L/2)$$10 + L/2 = 550 - 27.5L$$28L = 540$$L = 540 / 28 \approx 19.28 \ m$Rounding to $1$ decimal digit, $L = 19.3 \ m$.

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