Chapter 6, Problem 133P

(a)

To determine

The speed of the blocks moving along the inclines.

Answer to Problem 133P

The speed of the blocks is $1.7\text{m/s}$.

Explanation of Solution

Let us consider the distance be $d$ moved along the incline by mass $m_2$. Both masses move the same distance and have the same speed, since they are connected by a rope.

Write the equation for law of conservation of energy of the system.

$K_{1i}+K_{2i}+U_{1i}+U_{2i}=K_{1f}+K_{2f}+U_{1f}+U_{2f}$ (I)

Here, $K_{1i}$ is the initial kinetic energy of the block 1, $K_{2i}$ is the initial kinetic energy of the block 2, $U_{1i}$ is the initial potential energy due to gravity of the block 1, $U_{2i}$ is the initial potential energy due to gravity of the block 2, $K_{1f}$ is the final kinetic energy of the block 1, $K_{2f}$ is the final kinetic energy of the block 2, $U_{1f}$ is the final potential energy due to gravity of the block 1, $U_{2f}$ is the final potential energy due to gravity of the block 2,

Write the equation for final kinetic energy.

$K_{1f}=\frac{1}{2}{m}_1{v}^2$ (II)

Here, $m_1$ is the mass of the block 1 and $v$ is the velocity of the block.

Write the equation for final kinetic energy.

$K_{2f}=\frac{1}{2}{m}_2{v}^2$ (III)

Here, $m_2$ is the mass of the block 2 and $v$ is the velocity of the block.

Write the equation for initial potential energy.

$U_{1i}=m_{1}gd\sin \theta$ (IV)

Here, $g$ is the gravitational acceleration, $\theta$ is the angle makes with the block 2, and $d$ is the distance of the blocks travel.

Write the equation for final potential energy.

$U_{2f}=m_{2}gd\sin \varphi$ (V)

Here, $\varphi$ is the angle makes with the block 1.

Conclusion:

Substitute equation (II), (III), (IV) and (V) in equation (I).

$\begin{array}{c}0+0+m_{1}gd\sin \theta +0=\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2+0+m_{2}gd\sin \varphi \\ m_{1}gd\sin \theta =\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2+m_{2}gd\sin \varphi \\ m_{1}gd\sin \theta -m_{2}gd\sin \varphi =\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2\end{array}$

Solve the above equation for $v$.

$\begin{array}{c}gd\left(m_1\sin \theta -m_2\sin \varphi \right)=\frac{1}{2}{v}^2\left(m_1+m_2\right)\\ v^2=\frac{2gd\left(m_1\sin \theta -m_2\sin \varphi \right)}{\left(m_1+m_2\right)}\\ v=\sqrt{\frac{2gd\left(m_1\sin \theta -m_2\sin \varphi \right)}{\left(m_1+m_2\right)}}\end{array}$

Substitute $9.80{\text{m/s}}^2$ for $g$, $2.00\text{m}$ for $d$, $6.00\text{kg}$ for $m_1$, $36.9°$ for $\theta$, $45.0°$ for $\varphi$, and $4.00\text{kg}$ for $m_2$ in above equation.

$\begin{array}{c}v=\sqrt{\frac{2\left(9.80{\text{m/s}}^2\right)\left(2.00\text{m}\right)\left(6.00\text{kg}\sin 36.9°-4.00\text{kg}\sin 45.0°\right)}{\left(6.00\text{kg}+4.00\text{kg}\right)}}\\ =\sqrt{\frac{\left(39.2{\text{m}}^2/{\text{s}}^2\right)\left(3.60\text{kg}-2.83\text{kg}\right)}{\left(10.0\text{kg}\right)}}\\ =1.7\text{m/s}\end{array}$

Therefore, the speed of the blocks is $1.7\text{m/s}$.

(b)

To determine

Find the speed of block along the incline at constant acceleration.

Answer to Problem 133P

The speed of block along the incline at constant acceleration is $1.7\text{m/s}$.

Explanation of Solution

Write the equation for Newton’s second law applied for mass $m_1$.

$\begin{array}{c}{\displaystyle \sum F}=m_{1}a\\ m_{1}g\sin \theta -T=m_{1}a\end{array}$ (VI)

Here, $T$ is tension exerted on the block 1 and $a$ is the acceleration of the block.

Write the equation for Newton’s second law applied for mass $m_2$.

$\begin{array}{c}{\displaystyle \sum F}=m_{2}a\\ T-m_{2}g\sin \varphi =m_{2}a\end{array}$ (VII)

Rewrite the equation (VI) for tension force.

$T=m_{1}g\sin \theta -m_{1}a$ (VIII)

Rewrite the equation (VII) for tension force.

$T=m_{2}g\sin \varphi +m_{2}a$ (IX)

Compare the equation (VIII) and (IX) to solve acceleration.

$\begin{array}{c}{m}_{1}g\sin \theta -m_{1}a=m_{2}g\sin \varphi +m_{2}a\\ m_{1}g\sin \theta -m_{2}g\sin \varphi =m_{2}a+m_{1}a\\ g\left(m_1\sin \theta -m_2\sin \varphi \right)=a\left(m_1+m_2\right)\\ a=\frac{g\left(m_1\sin \theta -m_2\sin \varphi \right)}{\left(m_1+m_2\right)}\end{array}$ (X)

Write the equation for motion of blocks travel along the incline.

$v^2=u^2+2ad$ (XI)

Here, $v$ is the velocity of the block and $u$ is the initial velocity of the block.

Conclusion:

Substitute $9.80{\text{m/s}}^2$ for $g$, $2.00\text{m}$ for $d$, $6.00\text{kg}$ for $m_1$, $36.9°$ for $\theta$, $45.0°$ for $\varphi$, and $4.00\text{kg}$ for $m_2$ in equation (X).

$\begin{array}{c}a=\frac{9.80{\text{m/s}}^2\left(6.00\text{kg}\sin 36.9°-4.00\text{kg}\sin 45.0°\right)}{\left(6.00\text{kg}+4.00\text{kg}\right)}\\ =\frac{\left(9.80{\text{m/s}}^2\right)\left(3.60\text{kg}-2.83\text{kg}\right)}{\left(10.0\text{kg}\right)}\\ =0.76{\text{m/s}}^2\end{array}$

Susbtitute $0\text{m/s}$ for $u$, $0.76{\text{m/s}}^2$ for $a$, and $2.00\text{m}$ for $d$ in equation (XI)

$\begin{array}{c}{v}^2={\left(0\text{m/s}\right)}^2+2\left(0.76{\text{m/s}}^2\right)\left(2.00\text{m}\right)\\ v^2=3.04{\text{m}}^2/{\text{s}}^2\\ v=1.7\text{m/s}\end{array}$

Therefore, the speed of block along the incline at constant acceleration is $1.7\text{m/s}$.