Question

Consider two blocks pushed along a frictionless surface by someone who exerts a constant force with his or her hand (see the figure below). ©McGraw Hill What would be the magnitude of the force that the blocks exerted on each other if blocks A and B had masses of 10 kg and 7.5 kg and we pushed on the blocks with a force of 5 N?

Answer 1

In this case, the force that block A exerts on block B (and vice versa) is approximately 2.14 N.

Step-by-step explanation:

To find the magnitude of the force that blocks A and B exert on each other, we can use Newton's Second Law of motion. This law states that the force applied to an object is equal to the mass of the object times its acceleration (F = ma).

First, let's calculate the total mass of the system by adding the masses of block A and block B. Given that block A has a mass of 10 kg and block B has a mass of 7.5 kg, the total mass of the system is:

Total mass (m) = mass of block A + mass of block B = 10 kg + 7.5 kg = 17.5 kg

Next, we need to find the acceleration of the system. Since the blocks are on a frictionless surface and are being pushed together, they will accelerate as a single system. We can use Newton's Second Law to calculate the acceleration:

F = ma

Here, F is the total force applied to the system, which is 5 N, and m is the total mass of the system, which is 17.5 kg.

`a = F / m = 5 N / 17.5\ kg = 0.2857\ m/s\²`

Now that we have the acceleration, we can determine the force that block A exerts on block B (and vice versa), which is actually the same due to Newton's Third Law (every action has an equal and opposite reaction).

Let's consider block B with a mass of 7.5 kg. Using Newton's Second Law again:

`F_B = m_B * a`

Here,
`F_B`is the force exerted on block B,
`m_B` is the mass of block B (7.5 kg), and a is the acceleration we just calculated (0.2857 m/s²).

`F_B = 7.5 kg * 0.2857 m/s^2 = 2.14275 N`

Therefore, the force that block A exerts on block B (and vice versa) is approximately 2.14 N.