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## Q. 8.16

BLOCKS AND PULLEY

GOAL Solve a system requiring rotation concepts and the work-energy theorem.

PROBLEM Two blocks with masses $m_{1}=5.00 \mathrm{~kg}$ and $m_{2}=7.00 \mathrm{~kg}$ are attached by a string as in Figure 8.31a, over a pulley with mass $M=2.00 \mathrm{~kg}$. The pulley, which turns on a frictionless axle, is a hollow cylinder with radius $0.0500 \mathrm{~m}$ over which the string moves without slipping. The horizontal surface has coefficient of kinetic friction $0.350$. Find the speed of the system when the block of mass $m_{2}$ has dropped $2.00 \mathrm{~m}$.

STRATEGY This problem can be solved with the extension of the work-energy theorem, Equation 8.15b.

$KE_r = \frac{1}{2}I \omega^2$      [8.15]

If the block of mass $m_{2}$ falls from height $h$ to 0 , then the block of mass $m_{1}$ moves the same distance, $\Delta x=h$. Apply the work-energy theorem, solve for $v$, and substitute. Kinetic friction is the sole nonconservative force.

## Verified Solution

Apply the work-energy theorem, with $P E=P E_{g}$, the potential energy associated with gravity:

$W_{n c}=\Delta K E_{t}+\Delta K E_{r}+\Delta P E_{g}$

Substitute the frictional work for $W_{n c}$, kinetic energy changes for the two blocks, the rotational kinetic energy change for the pulley, and the potential energy change for the second block:

\begin{aligned}-\mu_{k} n \Delta x=&-\mu_{k}\left(m_{1} g\right) \Delta x=\left(\frac{1}{2} m_{1} v^{2}-0\right)+\left(\frac{1}{2} m_{2} v^{2}-0\right) \\&+\left(\begin{array}{ll}\frac{1}{2} I \omega^{2}-0\end{array}\right)+\left(\begin{array}{ll}0-m_{2} g h\end{array}\right)\end{aligned}

Substitute $\Delta x=h$, and write $I$ as $\left(I / r^{2}\right) r^{2}$ :

$-\mu_{k}\left(m_{1} g\right) h=\frac{1}{2} m_{1} v^{2}+\frac{1}{2} m_{2} v^{2}+\frac{1}{2}\left(\frac{I}{r^{2}}\right) r^{2} \omega^{2}-m_{2} g h$

For a hoop, $I=M r^{2}$ so $\left(I / r^{2}\right)=M$. Substitute this quantity and $v=r \omega$ :

$-\mu_{k}\left(m_{1} g\right) h=\frac{1}{2} m_{1} v^{2}+\frac{1}{2} m_{2} v^{2}+\frac{1}{2} M v^{2}-m_{2} g h$

Solve for $v$

\begin{aligned}m_{2} g h-\mu_{k}\left(m_{1} g\right) h &=\frac{1}{2} m_{1} v^{2}+\frac{1}{2} m_{2} v^{2}+\frac{1}{2} M v^{2} \\&=\frac{1}{2}\left(m_{1}+m_{2}+M\right) v^{2} \\v &=\sqrt{\frac{2 g h\left(m_{2}\right.}{\left.m_{1}+m_{2} m_{1}\right)}}\end{aligned}

Substitute $m_{1}=5.00 \mathrm{~kg}, m_{2}=7.00 \mathrm{~kg}, M=2.00 \mathrm{~kg}, g=9.80 \mathrm{~m} / \mathrm{s}^{2}, h=2.00 \mathrm{~m}$, and $\mu_{k}=0.350$ :

$v=3.83 \mathrm{~m} / \mathrm{s}$

REMARKS In the expression for the speed $v$, the mass $m_{1}$ of the first block and the mass $M$ of the pulley all appear in the denominator, reducing the speed, as they should. In the numerator, $m_{2}$ is positive while the friction term is negative. Both assertions are reasonable because the force of gravity on $m_{2}$ increases the speed of the system while the force of friction on $m_{1}$ slows it down. This problem can also be solved with Newton’s second law together with $\tau=I \alpha$, a good exercise.