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The problem involves a block of mass 6 kg on a smooth horizontal surface, with a 5 kg mass hanging on one side and a 3 kg mass hanging on the other side. We'll find:

(a) The acceleration of the entire system.
(b) The tension in the left-hand string.
(c) The tension in the right-hand string.

### Part (a): Finding the Acceleration of the Block
The total force acting on the system due to the hanging masses can be calculated by considering the difference in the gravitational forces acting on the 5 kg and 3 kg masses. The net force (\( F_{net} \)) causing the acceleration is due to the unbalanced weight of the two hanging masses.

The gravitational force acting on each mass is given by \( F = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

For the 5 kg mass:
\[ F_5 = 5 \times 9.81 = 49.05 \, \text{N} \]

For the 3 kg mass:
\[ F_3 = 3 \times 9.81 = 29.43 \, \text{N} \]

The net force (\( F_{net} \)) is the difference between these two forces, pulling the block towards the 5 kg side:
\[ F_{net} = F_5 - F_3 = 49.05 \, \text{N} - 29.43 \, \text{N} = 19.62 \, \text{N} \]

The total mass of the system that is being accelerated includes the 6 kg block and the masses of the hanging particles:
\[ M_{total} = 6 + 5 + 3 = 14 \, \text{kg} \]

Using Newton's second law (\( F = ma \)), the acceleration (\( a \)) of the system is:
\[ a = \frac{F_{net}}{M_{total}} = \frac{19.62}{14} \]

Let's calculate \( a \):

\[ a = \frac{19.62}{14} \approx 1.401 \, \text{m/s}^2 \]

### Part (b): Finding the Tension in the Left-Hand String
The tension in the left-hand string (T1) must balance the weight of the 5 kg mass while also accounting for its acceleration downward:
\[ T1 = F_5 - (5 \times a) \]

### Part (c): Finding the Tension in the Right-Hand String
Similarly, the tension in the right-hand string (T2) must balance the weight of the 3 kg mass while also accounting for its acceleration downward:
\[ T2 = F_3 - (3 \times a) \]

Let's compute the values for T1 and T2:
\[ T1 = 49.05 - (5 \times 1.401) \]
\[ T2 = 29.43 - (3 \times 1.401) \]

Let me calculate these tensions:The calculations yield the following results:

(a) The acceleration of the block is approximately \(1.401 \, \text{m/s}^2\).

(b) The tension in the left-hand string is approximately \(42.04 \, \text{N}\).

(c) The tension in the right-hand string is approximately \(25.23 \, \text{N}\).

Would you like further details on any part of this problem or have any other questions? Here are some related questions to consider:

1. What would be the effect on the acceleration if the surface was not frictionless?
2. How would the tensions change if the mass of the block was different?
3. What would happen if the masses on either side were equal?
4. How does the acceleration depend on the sum of the masses versus the difference in their weights?
5. What assumptions are critical for this solution to be valid?

**Tip:** When considering multiple bodies connected with strings and experiencing different forces, always ensure to consider the net force and total mass for calculating the system's acceleration.

The diagram shows a block of mass 6 kg, that is on a smooth horizontal surface. It is attached to particles of mass 5 kg and 3 kg. The problem asks to find (a) the acceleration of the block, (b) the tension in the left-hand string, and (c) the tension in the right-hand string.

Learn how to calculate the acceleration of a block on a smooth horizontal surface, with masses of 5kg and 3kg hanging on either side. This detailed step-by-step solution covers key physics concepts like Newton's Second Law and tension in strings, suitable for Grades 11-12 students.

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