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# An object is free to move on a table, except that there is a constant frictional force $f$ that opposes the motion of the ob- ject when it moves. If a force of $10.0$ N pulls the object, the acceleration is $2.0$ m/s$^2$. If a force of $20.0$ N pulls the object, the acceleration is $6.0$ m/s$^2$. $(a)$ What is the force of friction $f?$ $~~~~(A)$ $1.0$ N. $(B)$ $3.33$ N. $(C)$ $5.0$ N. $(D)$ $10.0$ N. $(b)$ What is the mass of the object? $~~~~(A)$ $0.40$ kg. $(B)$ $2.5$ kg. $(C)$ $3.33$ kg. $(D)$ $5.0$ kg.

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Calculate the friction force that opposes the motion of an object. It can be computed using the net force formula. Net force is the vector sum of forces acting object and friction opposes the motion of applied force to an object. We also know that net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Hence,

\begin{align*} F_{net} &= F_{app} - f \implies F_{net} = ma\\ \implies ma &= F_{app} - f \end{align*}

If a force of $10.0 \text{ N}$ pulls the object, $a = 2.0 \mathrm{~\dfrac{m}{s^2}}$. Entering these to equation, we obtain $m$ as

\begin{align*} m & = \dfrac{F_{app} - f}{a} = \dfrac{10.0 \text{ N} - f}{2.0 \mathrm{~\dfrac{m}{s^2}}} \end{align*}

If a force of $20.0 \text{ N}$ pulls the object, $a = 6.0 \mathrm{~\dfrac{m}{s^2}}$. Entering these to equation, we obtain $m$ as

\begin{align*} m & = \dfrac{20.0 \text{ N} - f}{6.0 \mathrm{~\dfrac{m}{s^2}}} \end{align*}

Substituting known values of $m$ gives

$\dfrac{10.0 \text{ N} - f}{2.0 \mathrm{~\dfrac{m}{s^2}}} = \dfrac{20.0 \text{ N} - f}{6.0 \mathrm{~\dfrac{m}{s^2}}}$

From this, we can solve for $f$:

\begin{align*} 30.0 \text{ N} - 3f & = 20.0 \text{ N} - f\\ f&=5.0 \text{ N} \end{align*}

$\boxed{\therefore \text{(C) } f=5.0 \text{ N} }$

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