Understanding the Physics Problem: Force and Acceleration

This problem involves applying Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this is represented as $\vec{F} = m\vec{a}$, where $\vec{F}$ is the net force, $m$ is the mass, and $\vec{a}$ is the acceleration.

We are given the same force acting on two different masses, resulting in different accelerations. We need to find the acceleration when these two masses are combined and the same force is applied.

Calculating Individual Masses using Newton's Second Law

We can use the given information for the individual blocks to find their masses, $m_1$ and $m_2$.

For the first block:

• Force ($F$) = 5 N
• Acceleration ($a_1$) = 10 m/s$^2$

Using Newton's second law, $F = m_1 a_1$:

$\qquad 5 \, \text{N} = m_1 \times 10 \, \text{m/s}^2$

Solving for $m_1$:

$\qquad m_1 = \frac{5 \, \text{N}}{10 \, \text{m/s}^2} = 0.5 \, \text{kg}$

For the second block:

• Force ($F$) = 5 N
• Acceleration ($a_2$) = 20 m/s$^2$

Using Newton's second law, $F = m_2 a_2$:

$\qquad 5 \, \text{N} = m_2 \times 20 \, \text{m/s}^2$

Solving for $m_2$:

$\qquad m_2 = \frac{5 \, \text{N}}{20 \, \text{m/s}^2} = 0.25 \, \text{kg}$

Finding the Combined Mass

When the two blocks are tied together, their masses combine. The total mass of the combined block, $m_{combined}$, is the sum of the individual masses:

$\qquad m_{combined} = m_1 + m_2$

$\qquad m_{combined} = 0.5 \, \text{kg} + 0.25 \, \text{kg} = 0.75 \, \text{kg}$

Calculating the Acceleration of the Combined System

Now, the same force of 5 N acts on the combined mass ($m_{combined} = 0.75$ kg). We can again use Newton's second law to find the acceleration ($a_{combined}$) of the combination:

$\qquad F = m_{combined} \times a_{combined}$

$\qquad 5 \, \text{N} = 0.75 \, \text{kg} \times a_{combined}$

Solving for $a_{combined}$:

$\qquad a_{combined} = \frac{5 \, \text{N}}{0.75 \, \text{kg}}$

To make the division easier, we can write 0.75 as a fraction, $\frac{3}{4}$:

$\qquad a_{combined} = \frac{5}{\frac{3}{4}} = 5 \times \frac{4}{3} = \frac{20}{3} \, \text{m/s}^2$

Converting the fraction to a decimal:

$\qquad a_{combined} \approx 6.666... \, \text{m/s}^2$

Rounding this to two decimal places, the acceleration is approximately 6.67 m/s$^2$.

Summary of Calculations

Comparing with Options

The calculated acceleration for the combined system is approximately 6.67 m/s$^2$. Let's look at the given options:

1. 6.67 m/s$^2$
2. 1.67 m/s$^2$
3. 5.67 m/s$^2$
4. 4.67 m/s$^2$

The calculated value matches the first option.

Revision Table: Key Concepts Revisited

Additional Information: Force, Mass, and Acceleration Relationship

Newton's second law highlights the fundamental relationship between force, mass, and acceleration. If the force is kept constant, the acceleration is inversely proportional to the mass. This means if you double the mass while applying the same force, the acceleration will be halved. Conversely, if you halve the mass, the acceleration will double.

In this problem, combining the two blocks results in a total mass ($0.75$ kg) that is greater than the mass of the second block ($0.25$ kg) but less than the mass of the first block ($0.5$ kg) - wait, let's recheck the masses: $m_1 = 0.5$ kg, $m_2 = 0.25$ kg. The combined mass is $0.75$ kg, which is greater than both individual masses. Therefore, when the same force is applied, the acceleration of the combined mass must be less than the acceleration of either individual mass. $a_1 = 10$ m/s$^2$, $a_2 = 20$ m/s$^2$. The combined acceleration is $6.67$ m/s$^2$, which is indeed less than both 10 and 20 m/s$^2$. This makes sense physically.

The formula $a = F/m$ shows that if F is constant, $a \propto 1/m$. Let the constant force be $F$. Then $m_1 = F/a_1$ and $m_2 = F/a_2$. The combined mass is $M = m_1 + m_2 = F/a_1 + F/a_2 = F(1/a_1 + 1/a_2) = F \frac{a_1 + a_2}{a_1 a_2}$. When force $F$ acts on mass $M$, the acceleration $a_{combined}$ is $F/M$.

$\qquad a_{combined} = \frac{F}{F \frac{a_1 + a_2}{a_1 a_2}} = \frac{1}{\frac{a_1 + a_2}{a_1 a_2}} = \frac{a_1 a_2}{a_1 + a_2}$

This formula is for the acceleration of combined mass when the same force acts on individual masses with accelerations $a_1$ and $a_2$. Let's check this with our values: $a_1 = 10$ m/s$^2$, $a_2 = 20$ m/s$^2$.

$\qquad a_{combined} = \frac{10 \times 20}{10 + 20} = \frac{200}{30} = \frac{20}{3} \approx 6.67 \, \text{m/s}^2$

This alternative method using inverse relationship also gives the same result, confirming our step-by-step calculation.