BLOCKS AND PULLEY GOAL Solve a system requiring rotation concepts and the work-energy theorem. PROBLEM Two blocks with masses m1=5.00 kg and m2=7.00 kg are attached by a string as in Figure 8.31a, over a pulley with mass M=2.00 kg. The pulley, which turns on a frictionless axle, is a hollow

Question

BLOCKS AND PULLEY

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

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

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

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

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

Accepted Answer

Apply the work-energy theorem, with [latex]P E=P E_{g}[/latex], the potential energy associated with gravity:

[latex]W_{n c}=\Delta K E_{t}+\Delta K E_{r}+\Delta P E_{g}[/latex]

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

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

Substitute [latex]\Delta x=h[/latex], and write [latex]I[/latex] as [latex]\left(I / r^{2} ight) r^{2}[/latex] :

[latex]-\mu_{k}\left(m_{1} g ight) h=\frac{1}{2} m_{1} v^{2}+\frac{1}{2} m_{2} v^{2}+\frac{1}{2}\left(\frac{I}{r^{2}} ight) r^{2} \omega^{2}-m_{2} g h[/latex]

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

[latex]-\mu_{k}\left(m_{1} g ight) 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[/latex]

Solve for [latex]v[/latex]

[latex]\begin{aligned}m_{2} g h-\mu_{k}\left(m_{1} g ight) 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 ight) v^{2} \v &=\sqrt{\frac{2 g h\left(m_{2} ight.}{\left.m_{1}+m_{2} m_{1} ight)}}\end{aligned}[/latex]

Substitute [latex]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}[/latex], and [latex]\mu_{k}=0.350[/latex] :

[latex]v=3.83 \mathrm{~m} / \mathrm{s}[/latex]

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

Question 8.16: BLOCKS AND PULLEY GOAL Solve a system requiring rotation con...

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