Physics word problem

Question

In Case A the mass of each block is m1 = m2 = 3.1 kg. In Case B the mass of block 1 (the block behind) is m1 = 6.2 kg, and the mass of

block 2 is m2 = 3.1 kg. No frictional force acts on block 1 in either Case A or Case B. However, a kinetic frictional force of 6.1 N does act

on block 2 in both cases and opposes the motion. For both Case A and Case B determine (a) the magnitude of the forces with which the blocks push against each other and (b) the magnitude of the acceleration of the blocks.

Answer 1

 Data:

  Friction = 0

  Kinetic Frictional Force on Block 2 = 6.1 N

Case A:

1. Mass of Blocks:

m1 =  m2​= 3.1kg

Case B:

1. Mass of Blocks:

m1=6.2 kg

m2=3.1 kg

Step 1: Net Force and Acceleration for Both Cases

Since both blocks are connected, they will experience the same acceleration. Let's first find the acceleration using Newton's second law.

Case 1:

Total mass =m1+m2=3.1 kg+3.1 kg=6.2 kg

Fnet​= −fk​= −6.1N

1. Using Newton's second law: f = m1a +m2.a

Case 2:

1. The total mass of the system:

Total mass =m1+m2=6.2 kg+3.1 kg=9.3 kg

The net external force acting on the system is still the kinetic frictional force on Block 2:

Fnet= −fk = −6.1 Using Newton's second law:

−6.1=9.3⋅a-6.1 = 9.3 \cdot a−6.1=9.3⋅a

Solving for acceleration a:

a ​= −0.656m/s2

Step 2: Force Between the Blocks

Now, let's find the force with which the blocks push against each other. This is the internal force between the two blocks.

Case 1:

1. The force on Block 2 due to Block 1 is simply the force required to accelerate Block 2:  

F = 3.05N (force between the blocks)

The negative sign indicates that the force is directed towards Block 1 (opposite to the direction of motion).

Case 2:

1. The force on Block 2 due to Block 1 :

F= −2.03N directed towards Block 1.

These are the forces with which the blocks push against each other and the accelerations for both cases.