Question

The coefficient of friction between the block of mass ${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.00}\mathbf{}\mathbf{}\mathit{k}\mathit{g}$and the surface in Figure P8.22 is.${\mathbf{\mu }}_k\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{400}$The system starts from rest. What is the speed of the ball ofmass ${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{.00}\mathbf{}\mathbf{}\mathit{k}\mathit{g}$whenit has fallen adistance$\mathbf{h}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{}\mathbf{m}$?

Short answer

The speed of the ball when it has fallen a distance $h=1.5\text{m}$ is$v=3.74\text{\hspace{0.17em}m/s}$ .

Step-by-step solution

Law of conservation of energy

In the presence of dissipative forces, the law of conservation of energy states that the total change in mechanical energy during a process, should be equal to the increase in internal energy of the system.

$\begin{array}{c}{\mathbf{\Delta E}}_{\mathrm{mech}}\mathbf{=}\mathbf{\Delta K}\mathbf{+}{\mathbf{\Delta U}}_g\\ \mathbf{=}\mathbf{-}{\mathbf{f}}_k\mathbf{d}\\ \mathbf{=}\mathbf{-}{\mathbf{\Delta E}}_{\mathrm{int}}...............(\mathbf{8}\mathbf{.}\mathbf{16})\end{array}$

If a system is not isolated as well, the total change in energy is given as-

$\begin{array}{c}\mathbf{\Delta E}\mathbf{=}\mathbf{\Delta K}\mathbf{+}\mathbf{\Delta U}\mathbf{+}{\mathbf{\Delta E}}_{\mathrm{int}}\\ \mathbf{=}\sum W_{\mathrm{other}\mathrm{forces}}\mathbf{.}\mathbf{...........}\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{17}\mathbf{)}\end{array}$

here,

$\mathbf{\Delta K}=$Change in kinetic energy

$\mathbf{\Delta U}=$Change in potential energy

${\mathbf{\Delta E}}_{\mathrm{int}}=$Change in internal energy

Given Data

mass of the block: $m_1=3.00\text{\hspace{0.17em}kg}$

mass of the ball: $m_2=5.00\text{\hspace{0.17em}kg}$

coefficient of friction: ${\mu }_k=0.400$

distance moved: $h=1.5\text{\hspace{0.17em}m}$

Step 3:

Use Equation 8.16:

$\begin{array}{c}\Delta E_{mech}=\Delta K+\Delta U_g\\ =-f_{k}d\\ =-\Delta E_{\mathrm{int}}\end{array}$

For the system of block, ball and surface; let the subscript ‘1’ and ‘2’ denote the values of block and ball respectively. So, equation becomes-

data-custom-editor="chemistry" $\Delta K_1+\Delta U_1+\Delta K_2+\Delta U_2=-f_{k}d$

Choose the initial point before release and the final point after each block have moved$1.5m$. Choose$u=0$ with the$3.00kg$ block on the tabletop and the block in its final position.

So $K_{i1}=K_{i2}=U_{1f}=U_{2f}=U_{i1}=0$

We now have

$\frac{1}{2}{m}_1{v}_f^2+\frac{1}{2}{m}_2{v}_f^2+0+0-0-0-0-m_{2}g{y}_{2i}=0-f_{k}d$

where the friction force is

$\begin{array}{c}{f}_k={\mu }_{k}n\\ ={\mu }_k{m}_{a}g\end{array}$

The friction force causes a negative change in mechanical energy because the force opposes the motion. Since all of the variables are known except for,$v_f$we can substitute and solve for the final speed.

$\frac{1}{2}{m}_1{v}_f^2+\frac{1}{2}{m}_2{v}_f^2-m_{2}g{y}_{2i}=-f_{k}d$

$\begin{array}{l}{v}^2=\frac{2gh(m_2-{\mu }_k{m}_1)}{m_1+m_2}\\ v=\sqrt{\frac{2\left({\text{9.8m/s}}^{\text{2}}\right)\left(\text{1.5m}\right)[5.\text{00kg}-\text{0.400}\left(\text{3.00kg}\right)]}{8.00\text{kg}}}\\ v=3.74\text{m/s}\end{array}$

After falling $1.5\text{\hspace{0.17em}m}$, the ball is moving with a velocity of .$3.74\text{m/s}$