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A 15Kg mass is accelerated from rest with a force of 100N. As it moves faster, friction and air resistance create an oppositely directed retarding force given by $F_{R} = A + B V$ . Where A=25 N and B= $0.5 \frac{N}{m / s}$ . At what velocity does the acceleration equal to one half of the initial acceleration ?

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$25 m s^{- 1}$

$50 m s^{- 1}$

$75 m s^{- 1}$

$100 m s^{- 1}$

answer is B.

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Detailed Solution

$F_{R} = A + B V = 25 + 0.5 V$

When $u = 0 F - F_{R} = m a$

$100 - 25 - 0.5 V_{0} = 15 a$

$a = 5 m s^{- 2}$ (initial acceleration)

$F - F_{R} = m a^{1}$ where $a^{1} = \frac{a}{2}$

$100 - 25 - 0.5 V = 15 \times \frac{5}{2}$

$\frac{75}{2} = 0.5 V V = 75 m s^{- 1}$

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