Chapter 4, Problem 121P

(a)

To determine

The acceleration of two blocks after they released from rest.

Answer to Problem 121P

Acceleration of block 1 is $3.9\text{m}/{\text{s}}^{\text{2}}$ in right direction.

Acceleration of block 2 is $3.9\text{m}/{\text{s}}^{\text{2}}$ downwards.

Explanation of Solution

Mass of block $m_1$ is $3.0\text{kg}$ and the mass of block $m_2$ is $2.0\text{kg}$.

The free body diagram is shown below.

Write the equation for net force on block 1 in vertical direction.

$N-m_{1}g=0$ (I)

Here, the normal reaction force on block 1 is $N$, mass of block 1 is $m_1$, and the gravitational acceleration is $g$.

Write the equation for net force on block 1 in horizontal direction.

$T-m_1{a}_{1x}=0$ (II)

Here, the tension on string is $T$, acceleration of block in horizontal direction is $a_{1x}$.

Write the equation for net force on block 2 in vertical direction.

$T-m_{2}g=m_2{a}_{2y}$ (III)

Here, the mass of block 2 is $m_2$ and the acceleration of $m_2$ in vertical direction is $a_{2y}$.

Write the condition to avoid the shrinking of cords.

$a_{1x}=-a_{2y}$

Here, the acceleration in right direction is taken as positive and in downward direction is taken as negative quantity.

Introduce a new common variable in place of $a_{1x}$ and $-a_{2y}$.

$a=a_{1x}$

Here, the new variable to denote the acceleration is $a$.

$a=-a_{2y}$

Rewrite equations

Rewrite equations (II) and (III) in terms of $T$ by substituting $a$ for $a_{1x}$ and $-a_{2y}$.

$\begin{array}{c}T=m_{1}a\\ T=m_{2}a+m_{2}g\end{array}$

Equate the right hand sides of above two equations.

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

Conclusion:

Substitute $3.0\text{kg}$ for $m_1$, $2.0\text{kg}$ for $m_2$, and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation to find $a$.

$\begin{array}{c}a=\frac{\left(2.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}{\left(3.0\text{kg}-2.0\text{kg}\right)}\\ =\frac{19.6\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{kg}}\\ =3.9\text{m}/{\text{s}}^{\text{2}}\end{array}$

The direction of acceleration of each block is same as that of the tension on string connecting the pulley and each block. Tension on string connecting $m_1$ and pulley is in right direction. Thus, acceleration of $m_1$ should be in right direction. Tension on string connecting $m_2$ and pulley is acting vertically downwards. Thus, acceleration of $m_2$ should be vertically downwards.

Therefore, the acceleration of block 1 is $3.9\text{m}/{\text{s}}^{\text{2}}$ in right direction and the acceleration of block 2 is $3.9\text{m}/{\text{s}}^{\text{2}}$ downwards.

(b)

To determine

The velocity of $m_1$ after $1.2\text{s}$ on the assumption that $m_1$ do not run out of room from the table and $m_2$ do not reaches the floor.

Answer to Problem 121P

The velocity is $4.7\text{m}/\text{s}$ in right direction.

Explanation of Solution

Mass of block $m_1$ is $3.0\text{kg}$ and the mass of block $m_2$ is $2.0\text{kg}$.

Write the Newton’s equation to find the velocity of $m_1$.

$v=u+a\Delta t$

Here, the velocity of $m_1$ is $v$, initial velocity is $u$, and the time taken is $\Delta t$.

Conclusion:

Substitute $0\text{m}/\text{s}$ for $u$, $3.9\text{m}/{\text{s}}^{\text{2}}$ for $a$, and $1.2\text{s}$ for $\Delta t$ in the above equation to find $v$.

$\begin{array}{c}v=0\text{m}/\text{s}+\left(3.9\text{m}/{\text{s}}^{\text{2}}\right)\left(1.2\text{s}\right)\\ =4.7\text{m}/\text{s}\end{array}$

Due to tension on string, $m_1$ moves in right direction. Thus, velocity is acting rightwards.

Therefore, the velocity is $4.7\text{m}/\text{s}$ in right direction.

(c)

To determine

The displacement of $m_1$ in $1.2\text{s}$.

Answer to Problem 121P

Displacement is $2.8\text{m}$ rightwards.

Explanation of Solution

Mass of block $m_1$ is $3.0\text{kg}$ and the mass of block $m_2$ is $2.0\text{kg}$.

Write the Newton’s equation to find the displacement of $m_1$.

$d=u\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2$ (IV)

Here, the displacement of $m_1$ is $d$, initial velocity is $u$, and the time taken is $\Delta t$.

Conclusion:

Substitute $0\text{m}/\text{s}$ for $u$, $3.9\text{m}/{\text{s}}^{\text{2}}$ for $a$, and $1.2\text{s}$ for $\Delta t$ in the above equation to find $d$.

$\begin{array}{c}d=\left(0\text{m}/\text{s}\right)\left(1.2\text{s}\right)+\frac{1}{2}\left(3.9\text{m}/{\text{s}}^{\text{2}}\right){\left(1.2\text{s}\right)}^2\\ =2.8\text{m}\end{array}$

$m_1$ is moving rightwards.

Therefore, the displacement is $2.8\text{m}$ rightwards.

(d)

To determine

The displacement of $m_1$ and $m_2$ at $0.40\text{s}$ after it is released from their initial positions.

Answer to Problem 121P

The displacement of $m_1$ is $0.31$ rightwards.

The displacement of $m_2$ is $0.31$ downwards.

Explanation of Solution

Mass of block $m_1$ is $3.0\text{kg}$ and the mass of block $m_2$ is $2.0\text{kg}$.

Conclusion:

Substitute $0\text{m}$ for $u$, $0.40\text{s}$ for $\Delta t$, and $3.9\text{m}/{\text{s}}^{\text{2}}$ for $a$ in equation (IV) to find $d$ for $m_1$.

$\begin{array}{c}d=\left(0\text{m}\right)\left(0.40\text{s}\right)+\frac{1}{2}\left(3.9\text{m}/{\text{s}}^{\text{2}}\right){\left(0.40\text{s}\right)}^2\\ =0.31\text{m}\end{array}$

Substitute $0\text{m}$ for $u$, $0.40\text{s}$ for $\Delta t$, and $3.9\text{m}/{\text{s}}^{\text{2}}$ for $a$ in equation (IV) to find $d$ for $m_1$.

$\begin{array}{c}d=\left(0\text{m}\right)\left(0.40\text{s}\right)+\frac{1}{2}\left(3.9\text{m}/{\text{s}}^{\text{2}}\right){\left(0.40\text{s}\right)}^2\\ =0.31\text{m}\end{array}$

$m_1$ is moving rightwards and $m_2$ moves vertically downwards.

Therefore, the displacement of $m_1$ is $0.31$ rightwards and the displacement of $m_2$ is $0.31$ downwards.