an object of mass $m$ falling near earths surface experiences a resistive force of $f_r=-bv^2$, where $b$ is…

an object of mass $m$ falling near earths surface experiences a resistive force of $f_r=-bv^2$, where $b$ is a positive constant with units of n $cdot$ s²/m² and $v$ is in m/s. what is the magnitude of the acceleration of the object when it is moving downward at a speed $v = sqrt{\frac{3mg}{4b}}$?\na $\frac{1}{4}g$\nb $\frac{3}{4}g$\nc $g$\nd $\frac{5}{4}g$

an object of mass $m$ falling near earths surface experiences a resistive force of $f_r=-bv^2$, where $b$ is a positive constant with units of n $cdot$ s²/m² and $v$ is in m/s. what is the magnitude of the acceleration of the object when it is moving downward at a speed $v = sqrt{\frac{3mg}{4b}}$?\na $\frac{1}{4}g$\nb $\frac{3}{4}g$\nc $g$\nd $\frac{5}{4}g$

Answer

Explanation:

Step1: Write the net - force equation

According to Newton's second law $F_{net}=ma$. The forces acting on the object are the gravitational force $F_g = mg$ (downward, positive direction) and the resistive force $F_R=-bv^2$ (upward, negative direction). So, $F_{net}=mg - bv^2=ma$.

Step2: Substitute the given value of $v$

We are given $v = \sqrt{\frac{3mg}{4b}}$. Substitute $v$ into the net - force equation: [ \begin{align*} mg - b\left(\sqrt{\frac{3mg}{4b}}\right)^2&=ma\ mg - b\times\frac{3mg}{4b}&=ma\ mg-\frac{3}{4}mg&=ma \end{align*} ]

Step3: Solve for $a$

Simplify the left - hand side of the equation: $\frac{4mg - 3mg}{4}=ma$, which gives $\frac{1}{4}mg=ma$. Divide both sides by $m$ (since $m\neq0$), we get $a=\frac{1}{4}g$.

Answer:

A. $\frac{1}{4}g$