Chapter 13, Problem 113A

(a)

To determine

To Calculate: The acceleration of the blocks.

Answer to Problem 113A

  $2.6{\text{m/s}}^{\text{2}}$

Explanation of Solution

Given:

Mass of hanging block, $m_1=3.0\text{kg}$

Mass of block on the plane, $m_2=2.0\text{kg}$

Formula used:

From Newton’s second law, the net force on the block causes it to accelerate.

  $F_{net}=ma$

Here, m is the mass and a is the acceleration

Calculation:

Let the tension in the string connecting the two blocks be T .

Free body diagram:

  

  $\begin{array}{l}{m}_{1}g-T=m_{1}a\mathrm{...}\left(1\right)\\ N=m_{2}g\cos \theta \\ T-m_{2}g\sin \theta -{\mu }_k{m}_{2}g\cos \theta =m_{2}a\mathrm{...}\left(2\right)\end{array}$

From (1)

  $m_{1}g-m_{1}a=T$

Substitute in (2) and solve

  $\begin{array}{l}{m}_{1}g-m_{1}a-m_{2}g\sin \theta -{\mu }_k{m}_{2}g\cos \theta =m_{2}a\\ m_{1}g-m_{2}g\sin \theta -{\mu }_k{m}_{2}g\cos \theta =\left(m_1+m_2\right)a\\ a=\frac{\left(m_1-m_2\sin \theta -{\mu }_k{m}_2\cos \theta \right)g}{\left(m_1+m_2\right)}\\ a=\frac{\left(3.0-2.0\sin 45°-0.19\left(2.0\right)\cos 45°\right)\left(9.8\right)}{\left(3.0+2.0\right)}\\ a=2.6{\text{m/s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the acceleration of the blocks is $2.6{\text{m/s}}^{\text{2}}$ .

(b)

To determine

To Calculate: The tension in the string.

Answer to Problem 113A

  $21.6\text{N}$

Explanation of Solution

Given:

Mass of hanging block, $m_1=3.0\text{kg}$

Mass of block on the plane, $m_2=2.0\text{kg}$

Formula used:

From Newton’s second law, the net force on the block causes it to accelerate.

  $F_{net}=ma$

Here, m is the mass and a is the acceleration

Calculation:

Let the tension in the string connecting the two blocks be T .

  $\begin{array}{l}{m}_{1}g-T=m_{1}a\mathrm{...}\left(1\right)\\ N=m_{2}g\cos \theta \\ T-m_{2}g\sin \theta -{\mu }_k{m}_{2}g\cos \theta =m_{2}a\mathrm{...}\left(2\right)\end{array}$

From (1)

  $m_{1}g-m_{1}a=T$

Substitute in value of acceleration in above equation and solve

  $\begin{array}{l}3.00\times 9.8-3.00\times 2.6{\text{m/s}}^{\text{2}}=T\\ T=21.6\text{N}\end{array}$

Conclusion:

Thus, the tension in the string is $21.6\text{N}$ .