A 15 kg mass is accelerated from rest with a force of 100 N. As it moves faster, friction and air resistance create an oppositively directed retrading 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?

25 m s^{- 1}

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50 m/s

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75 m/s

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100 m/s

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Solution

The correct option is C 75 m/s
Mass of the body $= 15$ kg
Initial velocity $= 0$ [since the body is accelerated from the rest position].
Force $= 100$ N.
In the Attachment it is clearly shown,
$100 - (25 + 0) = 15 a$ [since ,v is at rest intitially].
$\begin{matrix} \Rightarrow 15 a = 75 ∴ a = 5 m / s^{2} \end{matrix}$
Now, According to the question,
$\begin{matrix} \frac{5}{2} ˆ A = \frac{100 - (A + B v)}{15} \Rightarrow \frac{5}{2} = \frac{100 - (25 - 0.5 v)}{15} \Rightarrow 75 = 200 - 50 - v \Rightarrow 75 = 150 - v \Rightarrow v = 150 - 75 ∴ v = 75 m / s . \end{matrix}$
Hence, Option $C$ is correct answer.

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A 15 kg mass is accelerated from rest with a force of 100 N. As it moves faster, friction and air resistance create an oppositively directed retrading force given by FR=A+BV, Where A = 25 N and B = 0.5Nm/s. At what velocity does the acceleration equal to one half of the initial acceleration?