Newton’s Third Law and Interacting Objects – Step-by-Step Physics Guidance

Q1. What is the magnitude of the force that the 6.00-kg block exerts on the 4.00-kg block?

Background

Topic: Newton's Third Law & Newton's Second Law (Dynamics of Interacting Objects)

This question tests your understanding of how forces are transmitted between objects in contact, and how to apply Newton's laws to a system of connected masses on a frictionless surface.

Key Terms and Formulas

• Newton's Third Law: For every action, there is an equal and opposite reaction.

• Newton's Second Law: $F = ma$

• System: Both blocks together

• Internal force: Force that one block exerts on the other

Step-by-Step Guidance

1. First, consider both blocks as a single system. The total mass is $m_{\text{total}} = 6.00\ \text{kg} + 4.00\ \text{kg} = 10.00\ \text{kg}$.

2. Apply Newton's second law to the system: $F_{\text{net}} = m_{\text{total}} a$ where $F_{\text{net}}$ is the applied force (20.0 N).

3. Calculate the acceleration of the system: $a = \frac{F_{\text{net}}}{m_{\text{total}}}$

4. Now, focus on the 4.00-kg block. The only horizontal force acting on it is the force from the 6.00-kg block. This force causes the 4.00-kg block to accelerate at the same rate as the system.

5. Set up Newton's second law for the 4.00-kg block: $F_{6\rightarrow4} = m_{4} a$

Try solving on your own before revealing the answer!

Final Answer: 8.00 N

$a = \frac{20.0\ \text{N}}{10.0\ \text{kg}} = 2.00\ \text{m/s}^2$

$F_{6\rightarrow4} = 4.00\ \text{kg} \times 2.00\ \text{m/s}^2 = 8.00\ \text{N}$

The 6.00-kg block exerts a force of 8.00 N on the 4.00-kg block, which is what causes it to accelerate.

Q2. What is the acceleration of the 2.0-kg block in the three-object pulley system?

Background

Topic: Newton's Second Law, Tension, and Friction in Pulley Systems

This question tests your ability to analyze a system of masses connected by strings and pulleys, including the effects of friction.

Key Terms and Formulas

• Tension: The force transmitted by a string or rope.

• Kinetic friction: $f_k = \mu_k N$

• Newton's Second Law: $F = ma$

• Acceleration constraint: All connected objects accelerate together.

Step-by-Step Guidance

1. Draw free-body diagrams for each mass. Identify all forces: gravity, tension, and friction (for the 2.0-kg block).

2. Write Newton's second law for each block. For the 2.0-kg block: $T_1 - T_2 - f_k = m_2 a$

3. For the hanging blocks: $m_1 g - T_1 = m_1 a$ and $T_2 - m_3 g = m_3 a$

4. Express the friction force: $f_k = \mu_k m_2 g$

5. Combine the equations to solve for $a$ (the acceleration of the 2.0-kg block).

Try solving on your own before revealing the answer!

Final Answer: 1.7 m/s2

After combining the equations and substituting values, the acceleration is found to be 1.7 m/s2.

This result comes from balancing the net forces and accounting for friction.

Q3. What mass should block B have in order to start block A sliding up the plane?

Background

Topic: Static Friction, Inclined Planes, and Pulley Systems

This question tests your ability to analyze forces on an inclined plane and determine the minimum mass needed to overcome static friction.

Key Terms and Formulas

• Static friction: $f_s = \mu_s N$

• Newton's Second Law: $F = ma$ (here, $a = 0$ at the threshold)

• Inclined plane: $N = m_A g \cos\theta$

• Force up the plane: $T = m_B g$

Step-by-Step Guidance

1. Draw free-body diagrams for both blocks. Identify all forces: gravity, tension, normal force, and friction.

2. For block A, resolve forces parallel and perpendicular to the incline.

3. Set up the equation for the threshold of motion: $T = m_A g \sin\theta + f_s$

4. Express static friction: $f_s = \mu_s m_A g \cos\theta$

5. Set $T = m_B g$ and solve for $m_B$ in terms of the other quantities.

Try solving on your own before revealing the answer!

Final Answer: 2.6 kg

Plugging in the values and solving for $m_B$ gives the minimum mass needed to start block A sliding up the plane.

This ensures the tension just overcomes static friction and the component of gravity down the incline.

Q4. Two objects are connected by a very light flexible string as shown. (a) Draw free-body diagrams for each object. (b) Calculate the magnitude of the acceleration of each object. (c) Calculate the tension in the string.

Background

Topic: Atwood Machine, Newton's Second Law, Tension

This question tests your ability to analyze a classic Atwood machine setup, draw free-body diagrams, and calculate acceleration and tension.

Key Terms and Formulas

• Newton's Second Law: $F = ma$

• Tension: The force transmitted by the string

• Acceleration constraint: Both masses accelerate together

Step-by-Step Guidance

1. Draw free-body diagrams for both masses. For mass M: forces are gravity ($Mg$) and tension ($T$). For mass m: forces are gravity ($mg$) and tension ($T$).

2. Write Newton's second law for each mass: $M$ (down): $Mg - T = Ma$; $m$ (up): $T - mg = ma$

3. Add the two equations to eliminate $T$ and solve for $a$.

4. Once $a$ is found, substitute back to solve for $T$.

Try solving on your own before revealing the answer!

Final Answer: (a) Free-body diagrams as described. (b) $a = 2.0\ \text{m/s}^2$ (c) $T = 4.8\ \text{N}$

Using the equations, you find the acceleration and tension for the system.

The tension is less than the weight of the heavier mass, as expected for an accelerating system.

Q5. The figure shows a 100-kg block being released from rest from a height of 1.0 m. It then takes it 0.90 s to reach the floor. What is the mass m of the other block? The pulley has no appreciable mass or friction.

Background

Topic: Dynamics of Pulley Systems, Kinematics, Newton's Second Law

This question tests your ability to relate kinematics (distance, time) to dynamics (forces, masses) in a pulley system.

Key Terms and Formulas

• Kinematics: $d = \frac{1}{2} a t^2$ (for constant acceleration)

• Newton's Second Law: $F = ma$

• Forces: gravity, tension

• Acceleration constraint: Both masses accelerate together

Step-by-Step Guidance

1. Use the kinematic equation to solve for the acceleration: $d = \frac{1}{2} a t^2$

2. Write Newton's second law for both masses. For the 100-kg block: $100g - T = 100a$; for mass $m$: $T - mg = ma$

3. Add the equations to eliminate $T$ and solve for $m$ in terms of $a$ and $g$.

4. Substitute the value of $a$ found from the kinematics into the dynamics equation.

Try solving on your own before revealing the answer!

Final Answer: 54 kg

Solving the equations gives $m = 54\ \text{kg}$, which balances the system to produce the observed acceleration.

This ensures the 100-kg block falls the required distance in the given time.