Chapter 5, Problem 87A

a.

To determine

The acceleration of the blocks.

Answer to Problem 87A

  $3.96{\text{m/s}}^{\text{2}}$

Explanation of Solution

Given:

The mass of the block on the plane, $m_1=8.0\text{kg}$

The mass of the hanging block, $m_2=16.0\text{kg}$ .

The coefficient of the kinetic friction between the block and the plane, ${\mu }_k=0.23$

Formula used:

Newton’s Second law,

  $a=\frac{F_{net}}{\text{mass}}$

And, the kinetic friction acting on the object is given by,

  $f_k={\mu }_{k}N$

Where, ${\mu }_k$ is the coefficient of kinetic friction, and $N$ is the normal force acting on the object.

Calculation:

Let the hanging blockbe moving downwards with the acceleration $a$ , then the block on the plane will move up the plane with the same acceleration. Also, let the acceleration acting on the string be $T$ .

Now, the weight of the block on the plane will act in vertical downwards direction, dividing its weight into rectangular components as,

1^st component: Acting down the plane, parallel to it, $W_1=m_{1}g\sin \theta$

2^nd component: Acting perpendicular to the plane, in downwards direction, $W_2=m_{1}g\cos \theta$

Where $\theta =\text{ The angle of the incline}$

Since the block is not moving in the perpendicular direction of the plane, thus the normal force acting on the block will be balanced by the 2^nd component of weight, that is,

  $N=W_2=m_{1}g\cos \theta$

Then, the friction acting on the block will be,

  $f_k={\mu }_k{W}_2={\mu }_k{m}_{1}g\cos \theta$

Now, consider the motion of the block, up the incline. Then,

  $\begin{array}{l}T-\left(W_1+f_k\right)=m_{1}a\\ \Rightarrow T=m_{1}a+\left(m_{1}g\sin \theta +{\mu }_k{m}_{1}g\cos \theta \right)\mathrm{......}\left(1\right)\end{array}$

Now, consider the motion of the hanging block as,

  $m_{2}g-T=m_{2}a$

Using equation (1) in above,

  $\begin{array}{c}{m}_{2}g-m_{1}a-W_1-{\mu }_k{m}_{1}g\cos \theta =m_{2}a\\ a=\frac{m_{2}g-m_{1}g\sin \theta -{\mu }_k{m}_{1}g\cos \theta }{m_1+m_2}\end{array}$

Using given and known values in above,

  $\begin{array}{c}a=\frac{m_{2}g-m_{1}g\sin \theta -{\mu }_k{m}_{1}g\cos \theta }{m_1+m_2}\\ =\frac{\left(16\right)\left(9.8\right)-\left(\left(8\right)\left(9.8\right)\sin 37°\right)-\left(\left(0.23\right)\left(8\right)\left(9.8\right)\cos 37°\right)}{8+16}\\ =\frac{156.8-47.18-14.40}{24}\\ =3.96\text{}{\text{m/s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the acceleration of the block is $3.96{\text{m/s}}^{\text{2}}$ .

b.

To determine

The tension in the string connecting the blocks.

Answer to Problem 87A

  $93.26\text{N}$

Explanation of Solution

Given:

The mass of the block on the plane, $m_1=8.0\text{kg}$

The mass of the hanging block, $m_2=16.0\text{kg}$ .

The coefficient of the kinetic friction of the plane, ${\mu }_k=0.23$

From part (a), the acceleration of the block, $a=3.96{\text{m/s}}^{\text{2}}$

Formula used:

Newton’s Second law,

  $a=\frac{F_{net}}{\text{mass}}$

And, the kinetic friction acting on the object is given by,

  $f_k={\mu }_{k}N$

Where ${\mu }_k$ is the coefficient of kinetic friction, and $N$ is the normal force acting on the object.

Calculation:

From part (a), the Tension in the string is given in terms of $a$ as,

  $T=m_{1}a+\left(m_{1}g\sin \theta +{\mu }_k{m}_{1}g\cos \theta \right)\mathrm{......}\left(1\right)$

Then using the given and calculated values in above,

  $\begin{array}{c}T=8\left(3.96\right)+\left(\left(8\right)\left(9.8\right)\sin 37°+\left(0.23\right)\left(8\right)\left(9.8\right)\cos 37°\right)\\ =31.68+47.18+14.40\\ =93.26\text{}\text{N}\end{array}$

Conclusion:

Thus, the tension in the string connecting the blocks is $93.26\text{N}$ .