Chapter 4, Problem 87GP

(a)

To determine

To Find: Acceleration of each box.

Answer to Problem 87GP

  $a_1=4.05{\text{ m/s}}^2$ and $a_2=3.20{\text{ m/s}}^2$ .

Explanation of Solution

Given:

Mass of the box, $m_1=1.0\text{ kg}$

Coefficient of kinetic friction, ${\mu }_1=0.10$

Mass of the box, $m_2=2.0\text{ kg}$

Coefficient of kinetic friction, ${\mu }_2=0.20$

Angle of inclination, $\theta ={30}^0$

Calculation:

Net force on block with mass 1.0 kg is given by:

  $\begin{array}{l}{F}_1=m_{1}g\sin \theta -{\mu }_1{m}_{1}g\cos \theta \\ F_1=m_{1}g\left(\sin \theta -{\mu }_1\cos \theta \right)\end{array}$

Substituting the given values, we get:

  $\begin{array}{l}{F}_1=\left(1.0\text{ kg}\right)\left(9.8{\text{ m/s}}^2\right)\left(\sin {30}^0-\left(0.10\right)\cos {30}^0\right)\\ F_1=4.05\text{ N}\end{array}$

Hence, the acceleration is given by:

  $a_1=\frac{F_1}{m_1}=\frac{4.05\text{ N}}{1.0\text{ kg}}=4.05{\text{ m/s}}^2$

Similarly, the force on block with mass 2.0 kg is given by:

  $\begin{array}{l}{F}_2=m_{2}g\sin \theta -{\mu }_2{m}_{2}g\cos \theta \\ F_2=m_{2}g\left(\sin \theta -{\mu }_2\cos \theta \right)\\ F_2=\left(2.0\text{ kg}\right)\left(9.8{\text{ m/s}}^2\right)\left(\sin {30}^0-\left(0.20\right)\cos {30}^0\right)\\ F_2=6.40\text{ N}\\ a_2=\frac{F_2}{m_2}=\frac{6.40\text{ N}}{2.0\text{ kg}}=3.20{\text{ m/s}}^2\end{array}$

Conclusion:

Acceleration experienced by each box is: $a_1=4.05{\text{ m/s}}^2$ and $a_2=3.20{\text{ m/s}}^2$ .

(b)

To determine

The acceleration of each box if the taut spring is connected to the boxes.

Answer to Problem 87GP

  $a_1=4.05{\text{ m/s}}^2$ and $a_2=3.20{\text{ m/s}}^2$

Explanation of Solution

Given:

Mass of the box, $m_1=1.0\text{ kg}$

Coefficient of kinetic friction, ${\mu }_1=0.10$

Mass of the box, $m_2=2.0\text{ kg}$

Coefficient of kinetic friction, ${\mu }_2=0.20$

Angle of inclination, $\theta ={30}^0$

Calculation:

According to part a), the acceleration for each box is given by:

  $a_1=4.05{\text{ m/s}}^2$

  $a_2=3.20{\text{ m/s}}^2$

So, if $m_1$ has the acceleration greater than the acceleration of $m_2$ , then it can be concluded that the acceleration will be same as that of part a).

Conclusion:

Acceleration experienced by each box is: $a_1=4.05{\text{ m/s}}^2$ and $a_2=3.20{\text{ m/s}}^2$ .

(c)

To determine

To Find: The acceleration of each box, if the initial configuration is reversed.

Answer to Problem 87GP

The acceleration is $a=3.49{\text{ m/s}}^2$

Explanation of Solution

Given:

Mass of the box, $m_1=1.0\text{ kg}$

Coefficient of kinetic friction, ${\mu }_1=0.10$

Mass of the box, $m_2=2.0\text{ kg}$

Coefficient of kinetic friction, ${\mu }_2=0.20$

Angle of inclination, $\theta ={30}^0$

Calculation:

The net force can be given as:

  $\begin{array}{l}{F}_1+F_2=\left(4.05\text{ N}\right)+\left(6.40\text{ N}\right)\\ F=10.45\text{ N}\end{array}$

So, the acceleration can be calculated as follows:

  $\begin{array}{l}a=\frac{F}{m_1+m_2}\\ a=\frac{10.45\text{ N}}{1.0\text{ kg + 2}\text{.0 kg}}\\ a=3.49{\text{ m/s}}^2\end{array}$

Conclusion:

The acceleration is $a=3.49{\text{ m/s}}^2$