Question

Question: An object of mass $\mathit{m}$is sliding with speed ${\mathit{v}}_{\mathbf{i}}$at some instant across a level tabletop, with which its coefficient of kinetic friction is $\mathit{\mu }$. It then moves through a distance $\mathit{d}$and comes to rest. Which of the following equations for the speed ${\mathit{v}}_{\mathbf{i}}$is reasonable? (a) ${\mathit{v}}_{\mathbf{i}}\mathbf{=}\sqrt{\mathbf{-}\mathbf{2}\mathbf{\mu }\mathbf{m}\mathbf{g}\mathbf{d}}$ (b) ${\mathit{v}}_{\mathbf{i}}\mathbf{=}\sqrt{\mathbf{2}\mathbf{\mu mgd}}$ (c)${\mathit{v}}_{\mathbf{i}}\mathbf{=}\sqrt{\mathbf{-}\mathbf{2}\mathbf{\mu gd}}$ (d) ${\mathit{v}}_{\mathbf{i}}\mathbf{=}\sqrt{\mathbf{2}\mathbf{\mu gd}}$ (e) ${\mathit{v}}_{\mathbf{i}}\mathbf{=}\sqrt{\mathbf{2}\mathbf{\mu d}}$

Short answer

The correct option is (d): $v_{\mathrm{i}}=\sqrt{2\mathrm{\mu gd}}$.

Step-by-step solution

Write the given data from the question

The mass of the object is $m$.

The speed of the object is $v_i$.

The coefficient of kinetic friction is $\mu$.

The distance covered by the object is $d$.

The object comes to rest finally. Therefore, the final velocity of the object, $v=0$.

 Step 2: Determine the formulas to calculate the speed of the object

The kinetic equation of velocity is given as follows.

${\mathit{v}}^{\mathbf{2}}\mathbf{=}{\mathit{u}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{a}\mathit{s}$

Here, v is the final velocity, u is the initial velocity, s is the distance, and a is the acceleration.

The equation to calculate the frictional force is given as follows.

role="math" localid="1663734512150" ${\mathit{f}}_{\mathbf{k}}\mathbf{=}\mathit{\mu }\mathit{n}$

Here, n is the normal force.

The equation to calculate the force acting on the body is as follows.

role="math" localid="1663734499019" $\mathit{f}\mathbf{=}\mathit{m}\mathit{a}$

Calculate the velocity of the object

Consider the free-body diagram.

Consider the force in the direction.

$\begin{array}{l}\sum _{}{F}_y=0\\ n-m_g=0\\ n=m_g\end{array}$

Consider the force in the direction.

$$\begin{gathered} \sum _{}^{}{F}_x=0 \\ F+f_k=0 \\ F=-f_k \\ ma=-\mu n \end{gathered}$$

Substitute $mg$for $n$in the above equation.

$\begin{array}{l}ma=-\mu mg\\ a=-\mu g\end{array}$

Calculate the velocity of the object.

Substitute 0 for v, $-\mu g$for a, and $d$for $s$into equation (1).

$$\begin{gathered} 0=v_i^2+2\left(-\mu g\right)d \\ 0=v_i^2-2\mu gd \\ {v}_i^2=2\mu gd \\ {v}_i=\sqrt{2\mu gd} \end{gathered}$$

Hence, the speed of the object is $\sqrt{2\mu gd}$.