Math Problem Statement

Blocks with masses of 1.0 kg , 2.0 kg , and 3.0 kg are lined up in a row on a frictionless table. All three are pushed forward by a 12 N force applied to the 1.0 kg block. How much force does the 2.0 kg block exert on the 1.0 kg block?

Solution

Let's solve this problem step by step:

Given:

  • Mass of the 1.0 kg block: m1=1.0kgm_1 = 1.0 \, \text{kg}
  • Mass of the 2.0 kg block: m2=2.0kgm_2 = 2.0 \, \text{kg}
  • Mass of the 3.0 kg block: m3=3.0kgm_3 = 3.0 \, \text{kg}
  • Force applied: F=12NF = 12 \, \text{N}
  • The surface is frictionless, so there’s no friction.

Step 1: Find the acceleration of the system.

Since all three blocks are lined up and pushed by the same force, we can calculate the total mass of the system:

Total mass=m1+m2+m3=1.0kg+2.0kg+3.0kg=6.0kg\text{Total mass} = m_1 + m_2 + m_3 = 1.0 \, \text{kg} + 2.0 \, \text{kg} + 3.0 \, \text{kg} = 6.0 \, \text{kg}

Now, use Newton’s second law to find the acceleration aa of the entire system:

F=(m1+m2+m3)aF = (m_1 + m_2 + m_3) a

a=Fm1+m2+m3=12N6.0kg=2.0m/s2a = \frac{F}{m_1 + m_2 + m_3} = \frac{12 \, \text{N}}{6.0 \, \text{kg}} = 2.0 \, \text{m/s}^2

So, the acceleration of the entire system is a=2.0m/s2a = 2.0 \, \text{m/s}^2.

Step 2: Find the force exerted on the 1.0 kg block by the 2.0 kg block.

Now, we focus on the force exerted by the 2.0 kg block on the 1.0 kg block. Since the blocks are in contact, the force that the 2.0 kg block exerts on the 1.0 kg block is related to the acceleration of the 1.0 kg block.

The force on the 1.0 kg block (which is the force exerted by the 2.0 kg block) is:

F12=m1a=1.0kg×2.0m/s2=2.0NF_{1 \to 2} = m_1 a = 1.0 \, \text{kg} \times 2.0 \, \text{m/s}^2 = 2.0 \, \text{N}

Answer:

The 2.0 kg block exerts a force of 2.0 N on the 1.0 kg block.


Would you like to go over more details or have any further questions?

Here are some related questions:

  1. How much force does the 3.0 kg block exert on the 2.0 kg block?
  2. What would happen if there were friction on the table?
  3. How would the force on the 1.0 kg block change if the applied force were increased to 20 N?
  4. If the 2.0 kg block were removed, how would the acceleration of the system change?
  5. How would you calculate the net force on the 3.0 kg block in this system?

Tip: Remember that on a frictionless surface, all blocks will move with the same acceleration if they are in contact and pushed by a single external force.

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Math Problem Analysis

Mathematical Concepts

Newton's Laws of Motion
Force and Acceleration
Frictionless Motion

Formulas

F = ma
a = F / m
F = m1 * a

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 9-11