Question

Question: In Figure P5.92, the pulleys and the cord are light, all surfaces are frictionless, and the cord does not stretch. (a) How does the acceleration of block 1 compare with the acceleration of block 2 ? Explain your reasoning. (b) The mass of block 2 is 1.30 kg . Find its acceleration as it depends on the mass m₁ of block 1. (c) What If? What does the result of part (b) predict if m₁ is very much less than 1.30 kg ? (d) What does the result of part (b) predict if m₁ approaches infinity? (e) In this last case, what is the tension in the cord? (f) Could you anticipate the answers to parts (c), (d), and (e) without first doing part (b)? Explain.

Short answer

(a) The relation between the acceleration of the two blocks is $a_1=2a_2$.

(b) The acceleration of block 2 is $\frac{1300g}{\left(4m_1+1300\right)}\text{\hspace{0.33em}m}/\text{s}$.

(c) The acceleration of the block 2 is $9.8\text{\hspace{0.33em}m}/{\text{s}}^2$.

(d) The acceleration of the block 2 is 0.

(e) The tension in the chord is 6.37 N.

(f) Yes, the answer (c), (d) and (e) can be anticipate without doing the part (b).

Step-by-step solution

Writing the given data from the question

The mass of the block 2, m₂ = 1.30 kg = 1300 g.

The mass of the block 1 is very much less than 1.30 kg.

Determining the required formulas

The expression to calculate the force on the object is given as follows:

$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$

Here,F is the working acting on object, m is the mass of the object and, a is the acceleration of the object.

(a) Comparing the accelerations of the block 1 and block 2

Consider the free body diagram in which mass m₂ is moving in upward direction and mass m₁is moving in the right direction.

Consider the separate free body diagram of the two masses.

Since the two masses are connected with the different chords, therefore the acceleration of the both the masses would be different. The block m₂ is connected with the two chords and the both the ropes of the block 2 lengthened by the one-half meter.

The travel time of both the masses are same but distance is double in case of mass 2. Therefore, the acceleration of the block 2 is half of the acceleration of block 1.

$\begin{array}{l}{a}_2=\frac{a_1}{2}\\ a_1=2a_2\end{array}$

Hence, the relation between the acceleration of the two blocks is .$a_1=2a_2$

(b) Calculate the acceleration of block 2 in terms of the mass m1 of block 1 :

Consider the equilibrium in the horizontal direction for block m₁.

$\begin{array}{c}\sum F_x=T\\ m_1{a}_1=T\end{array}$

Consider the equilibrium in the vertical direction for block m₂.

role="math" localid="1663594274405" $$\begin{gathered} \sum F_y=m_{2}g-2T \\ {m}_2{a}_2=m_{2}g-2T \end{gathered}$$

Substitute a₁ / 2 for a₂ into the above equation.

$m_2\frac{a_1}{2}=m_{2}g-2T$

Substitute m₁a₁ for T into the above equation.

role="math" localid="1663594372739" $$\begin{gathered} \begin{array}{l}{m}_2\frac{a_1}{2}=m_{2}g-2m_1{a}_1\\ 2m_1{a}_1+\frac{m_2{a}_1}{2}=m_{2}g\\ a_1\left(2m_1+\frac{m_2}{2}\right)=m_{2}g\end{array} \\ {a}_1=\frac{m_{2}g}{\left(2m_1+\frac{m_2}{2}\right)} \end{gathered}$$

Calculate the acceleration of the block 2.

$a_2=\frac{a_1}{2}$

Substitute $\frac{m_{2}g}{\left(2m_1+\frac{m_2}{2}\right)}$ for a₁ into the above equation.

$$\begin{gathered} a_2=\frac{m_{2}g}{2\left(2m_1+\frac{m_2}{2}\right)} \\ {a}_2=\frac{m_{2}g}{\left(4m_1+m_2\right)}.......................................\left(ii\right) \end{gathered}$$

Substitute 1300 g for m₂ into the above equation.

$a_2=\frac{1300\text{g}}{\left(4m_1+1300\text{g}\right)}\text{\hspace{0.33em}m}/{\text{s}}^2$

Hence, the acceleration of block 2 is $\frac{1300\text{g}}{\left(4m_1+1300\text{g}\right)}\text{\hspace{0.33em}m}/{\text{s}}^2$.

(c) Calculating the acceleration of the block 2

Recall equation (ii).

$a_2=\frac{m_{2}g}{\left(4m_1+m_2\right)}$

Since the mass of the block 1 is very much less than block 2, therefore $4m_1<<<m_2$.

$\begin{array}{c}{a}_2=\frac{m_{2}g}{\left(m_2\right)}\\ =g\\ =9.8\text{\hspace{0.33em}m}/{\text{s}}^2\end{array}$

Hence, the acceleration of the block 2 is 9.8 m/s².

(d) Calculating the acceleration of the block 2, if the mass of block 1 approaches to infinity

The mass of block 1 approaches to infinity, $m_1=\infty$

Recall equation (ii).

$a_2=\frac{m_{2}g}{\left(4m_1+m_2\right)}$

Substitute $\infty$ for m₁ into the above equation.

$\begin{array}{c}{a}_2=\frac{m_{2}g}{4\left(\infty \right)+m_2}\\ =\frac{m_{2}g}{\infty }\\ =0\end{array}$

Hence, the acceleration of the block 2 is 0.

(e) Calculating the tension in the chords, if the mass of block 1 approaches infinity

The acceleration of the block 2 is zero when mass of block is approaches to infinity.

Recall the equation (i).

$m_2{a}_2=m_{2}g-2T$

Substitute 0 for a₂ into the above equation.

$\begin{array}{c}{m}_2\left(0\right)=m_{2}g-2T\\ 0=m_{2}g-2T\\ m_{2}g=2T\\ T=\frac{m_{2}g}{2}\end{array}$

Substitute 1.3 kg for m₂ and 9.8 m/s² for g into the above equation.

$\begin{array}{c}T=\frac{\left(1.3\text{\hspace{0.33em}kg}\right)\times \left(9.8\text{\hspace{0.33em}m}/{\text{s}}^2\right)}{2}\\ =\left(13\times 4.9\right)\text{N}\\ =6.37\text{N}\end{array}$

Hence, the tension in the chord is 6.37 N.

(f) Determining if the answers to parts (c), (d), and (e) without first doing part (b):

Yes, the answer (c), (d) and (e) can be anticipate without doing the part (b).

In part (c), the acceleration of the block 1 is free fall, therefore acceleration of block 1 is 9.8 m/s² . We have determined from the part (a) that $a_2=\frac{a_1}{2}$ therefore, acceleration of the block 2 is $4.9\text{\hspace{0.33em}m}/{\text{s}}^2$.

In part (d), the mass of block 1 is approaches to infinity, So, acceleration of the block 2 is zero. In part (e), the mass of block 1 is approaches to infinity, so the tension in chord of block 1 is zero. But the tension in the chord of the block is $6.37\text{N}$.