Question

Shows two blocks, masses 4.3 kg and 5.4 kg, being pushed across a frictionless surface by a 22.5-N horizontal force applied to the 4.3-kg block. A. What is the acceleration of the blocks? B. What is the force of the 4.3-kg block on the 5.4 -kg block? C. What is the force of the 5.4 -kg block on the 4.3 -kg block?

126

likes629 views

Gene

4.5

57 Answers

A. To find the acceleration of the blocks, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

Since the surface is frictionless, the only horizontal force acting on the system is the 22.5-N force applied to the 4.3-kg block. Therefore, the net force acting on the system is 22.5 N.

Let's denote the acceleration of the blocks asa and the mass of the 4.3-kg block as m_1 , and the mass of the 5.4-kg block as m_2 .

Using Newton's second law, we have:

F_{\text{net}} = (m_1 + m_2) \cdot a

Substituting the given values, we have:

22.5 \, \text{N} = (4.3 \, \text{kg} + 5.4 \, \text{kg}) \cdot a

Simplifying, we get:

22.5 \, \text{N} = 9.7 \, \text{kg} \cdot a

To find the acceleration, we can divide both sides of the equation by 9.7 kg:

a = \frac{22.5 \, \text{N}}{9.7 \, \text{kg}}

Calculating, we find:

a \approx 2.32 \, \text{m/s}^2

Therefore, the acceleration of the blocks is approximately2.32 \, \text{m/s}^2 .

Answer: a \approx 2.32 \, \text{m/s}^2

B. To find the force of the 4.3-kg block on the 5.4-kg block, we can use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.

Since the 4.3-kg block exerts a force on the 5.4-kg block, the 5.4-kg block exerts an equal but opposite force on the 4.3-kg block.

Therefore, the force of the 4.3-kg block on the 5.4-kg block is the same as the force of the 5.4-kg block on the 4.3-kg block.

Answer: The force of the 4.3-kg block on the 5.4-kg block is equal to the force of the 5.4-kg block on the 4.3-kg block.

C. The force of the 5.4-kg block on the 4.3-kg block is the same as the force of the 4.3-kg block on the 5.4-kg block, due to Newton's third law of motion.

Answer: The force of the 5.4-kg block on the 4.3-kg block is equal to the force of the 4.3-kg block on the 5.4-kg block.

Since the surface is frictionless, the only horizontal force acting on the system is the 22.5-N force applied to the 4.3-kg block. Therefore, the net force acting on the system is 22.5 N.

Let's denote the acceleration of the blocks as

Using Newton's second law, we have:

Substituting the given values, we have:

Simplifying, we get:

To find the acceleration, we can divide both sides of the equation by 9.7 kg:

Calculating, we find:

Therefore, the acceleration of the blocks is approximately

Answer:

B. To find the force of the 4.3-kg block on the 5.4-kg block, we can use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.

Since the 4.3-kg block exerts a force on the 5.4-kg block, the 5.4-kg block exerts an equal but opposite force on the 4.3-kg block.

Therefore, the force of the 4.3-kg block on the 5.4-kg block is the same as the force of the 5.4-kg block on the 4.3-kg block.

Answer: The force of the 4.3-kg block on the 5.4-kg block is equal to the force of the 5.4-kg block on the 4.3-kg block.

C. The force of the 5.4-kg block on the 4.3-kg block is the same as the force of the 4.3-kg block on the 5.4-kg block, due to Newton's third law of motion.

Answer: The force of the 5.4-kg block on the 4.3-kg block is equal to the force of the 4.3-kg block on the 5.4-kg block.

Frequently asked questions (FAQs)

What is the average age (in years) of the students in a class, given that there are 20 students aged 12, 15 students aged 13, and 10 students aged 14?

+

What is the common radian measure of an angle subtending an arc of length 5 cm on a unit circle?

+

What are the components of a unit vector projected onto the xy-plane, knowing the angle it makes with the positive x-axis?

+

New questions in Mathematics