Chapter 5, Problem 90AP

(a)

To determine

The acceleration with which each blocks starts to move.

Answer to Problem 90AP

Solution: The acceleration of the left block is $-1.23{\text{m/s}}^{\text{2}}$.

The acceleration of the right block is $0.616{\text{m/s}}^{\text{2}}$.

Explanation of Solution

Given Info: The force constant of the spring is $3.85\text{N/m}$, the length the spring is compressed by the blocks is $8.0\text{cm}$, the mass of the block in the left side is $0.250\text{kg}$, the mass of the block in the right side of the spring is $0.500\text{kg}$, the coefficient of kinetic friction is zero.

Since, the coefficient of kinetic friction is zero; $f_{r,1}=f_{r,2}=0$.

Consider right side as positive.

The expression used to calculate the acceleration of block.

$a_1=\frac{-F}{m_1}$

• $F$ is the spring force exerted on the block
• $a_1$ is the acceleration of the left block
• $m_1$ is the mass of the left block

Re-write the expression.

$a_1=\frac{kx}{m_1}$

• k is the force constant of the spring

Substitute $3.85\text{N/m}$ for k, $8.0\text{cm}$ for x and $0.250\text{kg}$ for $m_1$ to find the acceleration of the left block.

$\begin{array}{c}{a}_1=\frac{-\left(3.85\text{N/m}\right)\left(8.0\times {10}^{-2}\text{m}\right)}{0.250\text{kg}}\\ =\frac{-0.308\text{N}}{0.250\text{kg}}\\ =-1.23{\text{m/s}}^{\text{2}}\end{array}$

Since, the coefficient of kinetic friction is zero; $f_{r,1}=f_{r,2}=0$.

Consider right side as positive.

The expression used to calculate the acceleration of block.

$a_2=\frac{-F}{m_2}$

• $F$ is the spring force exerted on the block
• $a_2$ is the acceleration of the left block
• $m_2$ is the mass of the left block

Re-write the expression.

$a_2=\frac{kx}{m_2}$

• k is the force constant of the spring

Substitute $3.85\text{N/m}$ for k, $8.0\text{cm}$ forx and $0.500\text{kg}$ for $m_2$ to find the acceleration of the right block.

$\begin{array}{c}{a}_2=\frac{\left(3.85\text{N/m}\right)\left(8.0\times {10}^{-2}\text{m}\right)}{0.500\text{kg}}\\ =\frac{0.308\text{N}}{0.500\text{kg}}\\ =0.616{\text{m/s}}^{\text{2}}\end{array}$

Thus, the acceleration of the right block is $0.616{\text{m/s}}^{\text{2}}$.

Conclusion:

Therefore, the acceleration of the left block is $-1.23{\text{m/s}}^{\text{2}}$ and  the acceleration of the right block is $0.616{\text{m/s}}^{\text{2}}$.

(b)

To determine

The acceleration with which each blocks starts to move.

Answer to Problem 90AP

Solution: the acceleration of the left block is $-0.252{\text{m/s}}^{\text{2}}$.

the acceleration of right block is zero.

Explanation of Solution

Given Info: The force constant of the spring is $3.85\text{N/m}$, the length the spring is compressed by the blocks is $8.0\text{cm}$, the mass of the block in the left side is $0.250\text{kg}$, the mass of the block in the right side of the spring is $0.500\text{kg}$, the coefficient of kinetic friction is $0.100$.

Consider right side as positive.

The expression used to calculate the acceleration of block.

$a_1=\frac{-F+f_{k,1}}{m_1}$

• $F$ is the spring force exerted on the block
• $a_1$ is the acceleration of the left block
• $m_1$ is the mass of the left block
• $f_{k,1}$ is kinetic friction force

Re-write the expression.

$\begin{array}{c}{a}_1=\frac{kx+{\mu }_{k}n}{m_1}\\ =\frac{kx+{\mu }_k\left(m_{1}g\right)}{m_1}\end{array}$

• k is the force constant of the spring
• ${\mu }_k$ is the coefficient of kinetic friction
• n is the normal force

Substitute $3.85\text{N/m}$ for k, $8.0\text{cm}$ for x, $0.250\text{kg}$ for $m_1$, $0.100$ for ${\mu }_k$ and $9.\text{8}{\text{m/s}}^{\text{2}}$ forg to find the acceleration of the left block.

$\begin{array}{c}{a}_1=\frac{-\left(3.85\text{N/m}\right)+\left(0.250\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)}{0.250\text{kg}}\\ =\frac{-\left(3.85\text{N/m}\right)+\left(0.245\text{N}\right)}{0.250\text{kg}}\\ =-0.252{\text{m/s}}^{\text{2}}\end{array}$

Thus, the acceleration of the left block is $-0.252{\text{m/s}}^{\text{2}}$.

The expression used to calculate the kinetic friction force.

$\begin{array}{c}{f}_{k,2}={\mu }_{k}n\\ ={\mu }_k\left(m_{2}g\right)\end{array}$

Substitute $0.100$ for ${\mu }_k$, $0.500\text{kg}$ for $m_2$ and $9.8{\text{m/s}}^{\text{2}}$ for g to find the kinetic frictional force for right block.

$\begin{array}{c}{f}_{k,2}=\left(0.100\right)\left(0.500\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ =0.490\text{N}\end{array}$

Since, the kinetic frictional force, $0.490\text{N}$ is more than the spring force, $0.308\text{N}$ exerted on the block; the acceleration of the right block is zero.

Thus, $a_2=0$.

Conclusion:

Therefore, the acceleration of the left block is $-0.252{\text{m/s}}^{\text{2}}$ and the acceleration of right block is zero.

(c)

To determine

The acceleration with which both blocks start to move.

Answer to Problem 90AP

Solution: The acceleration with which both blocks start to move is zero.

Explanation of Solution

Given Info: The force constant of the spring is $3.85\text{N/m}$, the length the spring is compressed by the blocks is $8.0\text{cm}$, the mass of the block in the right side is $0.500\text{kg}$, the mass of the block in the left side is $0.250\text{kg}$, the coefficient of kinetic friction is $0.462$.

The expression used to calculate the kinetic friction force.

$\begin{array}{c}{f}_{k,2}={\mu }_{k}n\\ ={\mu }_k\left(m_{2}g\right)\end{array}$

Substitute $0.462$ for ${\mu }_k$, $0.500\text{kg}$ for $m_2$ and $9.8{\text{m/s}}^{\text{2}}$ for g to find the kinetic frictional force for right block.

$\begin{array}{c}{f}_{k,2}=\left(0.462\right)\left(0.500\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ =2.26\text{N}\end{array}$

Since, the kinetic frictional force, $2.26\text{N}$ is more than the spring force, $0.308\text{N}$ exerted on the block; the acceleration of the right block is zero.

The expression used to calculate the kinetic friction force.

$\begin{array}{c}{f}_{k,1}={\mu }_{k}n\\ ={\mu }_k\left(m_{1}g\right)\end{array}$

Substitute $0.462$ for ${\mu }_k$, $0.250\text{kg}$ for $m_1$ and $9.8{\text{m/s}}^{\text{2}}$ for g to find the kinetic frictional force for left block.

$\begin{array}{c}{f}_{k,2}=\left(0.462\right)\left(0.250\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ =1.13\text{N}\end{array}$

Since, the kinetic frictional force, $1.13\text{N}$ is more than the spring force, $0.308\text{N}$ exerted on the block; the acceleration of the left block is zero.

Thus, $a_1=0$.

Conclusion

The acceleration with which both blocks start to move is zero