In Figure P4.53, the coefficient of kinetic friction between the two blocks shown is 0.26. The surface of the table and the pulleys are frictionless.Figure P4.53(a) Draw a free-body diagram for each block. (Do this on paper. Your instructor may ask you to turn in this work.)(b) Determine the magnitude of the acceleration of each block. 2 kg block 122.908Your response differs from the correct answer by more than 100%. m/s2 3 kg block 2Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2 10 kg block 3Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2 (c) Find the tension in the strings. T1 4Your response differs from the correct answer by more than 10%. Double check your calculations. N T2 5 N Hint: Active Figure 4.18

To tackle this problem, we need to analyze the forces acting on each block and apply Newton's second law of motion. Let's break it down step by step, focusing on the free-body diagrams, the calculations for acceleration, and the tension in the strings.

Free-Body Diagrams

Before we dive into calculations, it's essential to draw free-body diagrams for each block. This visual representation helps us identify all the forces acting on the blocks.

• For the 2 kg block: The forces acting on it include the gravitational force (weight) downward, which is \( F_g = m \cdot g = 2 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 19.62 \, \text{N} \), and the tension \( T_1 \) acting upward. The frictional force opposing the motion can be calculated as \( F_f = \mu_k \cdot F_n \), where \( F_n \) is the normal force, which equals the weight of the block in this case.
• For the 3 kg block: Similar to the 2 kg block, it experiences a downward gravitational force of \( 29.43 \, \text{N} \) and an upward tension \( T_2 \). The frictional force will also be calculated using the coefficient of kinetic friction.
• For the 10 kg block: This block will have a gravitational force of \( 98.1 \, \text{N} \) acting downward and tension \( T_2 \) acting upward.

Calculating Acceleration

Now that we have the free-body diagrams, we can set up the equations of motion for each block. Let's denote the acceleration of the system as \( a \).

For the 2 kg Block

The net force acting on the 2 kg block can be expressed as:

Net Force = T_1 - F_f

Where \( F_f = \mu_k \cdot F_n = 0.26 \cdot 19.62 \, \text{N} \approx 5.1 \, \text{N} \).

Thus, the equation becomes:

2a = T_1 - 5.1

For the 3 kg Block

The equation for the 3 kg block is:

Net Force = T_2 - F_f

Here, \( F_f = 0.26 \cdot 29.43 \, \text{N} \approx 7.66 \, \text{N} \).

So, we have:

3a = T_2 - 7.66

For the 10 kg Block

The equation for the 10 kg block is:

Net Force = F_g - T_2

Where \( F_g = 98.1 \, \text{N} \), leading to:

10a = 98.1 - T_2

Solving the System of Equations

Now we have three equations:

• 1) \( 2a = T_1 - 5.1 \)
• 2) \( 3a = T_2 - 7.66 \)
• 3) \( 10a = 98.1 - T_2 \)

We can express \( T_1 \) and \( T_2 \) in terms of \( a \) from the first two equations:

T_1 = 2a + 5.1

T_2 = 3a + 7.66

Substituting \( T_2 \) into the third equation gives:

10a = 98.1 - (3a + 7.66)

Solving this leads to:

10a = 98.1 - 3a - 7.66

13a = 90.44

a ≈ 6.96 \, \text{m/s}^2

Finding Tension in the Strings

Now that we have the acceleration, we can find the tensions:

Substituting \( a \) back into the equations for \( T_1 \) and \( T_2 \):

T_1 = 2(6.96) + 5.1 ≈ 18.02 \, \text{N}

T_2 = 3(6.96) + 7.66 ≈ 27.54 \, \text{N}

In summary, the acceleration of the blocks is approximately \( 6.96 \, \text{m/s}^2 \), with tensions \( T_1 \) and \( T_2 \) being approximately \( 18.02 \, \text{N} \) and \( 27.54 \, \text{N} \), respectively. Make sure to double-check your calculations and ensure that all units are consistent throughout the process!