## Chapter 8

## Q. 8.16

## 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.

## Step-by-Step

## 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.