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The problem describes a system with two blocks connected by a rope passing over a pulley. One block with a weight of \( 422 \, \text{N} \) is on a table, while the other block with a weight of \( 185 \, \text{N} \) hangs vertically. We are tasked with finding:

(a) The acceleration of both blocks.
(b) The tension in the rope.

### Given:
- Weight of the block on the table: \( W_1 = 422 \, \text{N} \)
- Weight of the hanging block: \( W_2 = 185 \, \text{N} \)
- Gravitational acceleration: \( g = 9.8 \, \text{m/s}^2 \)
- Ignore friction.

### Step 1: Find the masses of both blocks
The mass \( m \) of an object can be calculated from its weight \( W \) using the relation:
\[
m = \frac{W}{g}
\]
For the block on the table (\( m_1 \)):
\[
m_1 = \frac{422 \, \text{N}}{9.8 \, \text{m/s}^2} = 43.06 \, \text{kg}
\]

For the hanging block (\( m_2 \)):
\[
m_2 = \frac{185 \, \text{N}}{9.8 \, \text{m/s}^2} = 18.88 \, \text{kg}
\]

### Step 2: Set up the equations for the system
Let \( T \) represent the tension in the rope and \( a \) the acceleration of both blocks. Since the system is connected by a rope, the blocks will share the same acceleration magnitude, but in opposite directions.

For the block on the table (\( m_1 \)):
\[
T = m_1 a
\]
For the hanging block (\( m_2 \)), the forces acting are the tension upward and the weight downward. Therefore, applying Newton's second law:
\[
m_2 g - T = m_2 a
\]

### Step 3: Solve the system of equations
We now have two equations:
1. \( T = m_1 a \)
2. \( m_2 g - T = m_2 a \)

Substitute \( T = m_1 a \) into the second equation:
\[
m_2 g - m_1 a = m_2 a
\]

Rearrange to solve for acceleration \( a \):
\[
m_2 g = (m_1 + m_2) a
\]
\[
a = \frac{m_2 g}{m_1 + m_2}
\]
Substitute the known values:
\[
a = \frac{18.88 \times 9.8}{43.06 + 18.88} = \frac{184.22}{61.94} = 2.97 \, \text{m/s}^2
\]

### Step 4: Find the tension in the rope
Now that we have the acceleration, substitute it back into the equation \( T = m_1 a \):
\[
T = 43.06 \times 2.97 = 127.98 \, \text{N}
\]

### Final Answers:
(a) The acceleration of both blocks is \( 2.97 \, \text{m/s}^2 \).  
(b) The tension in the rope is \( 127.98 \, \text{N} \).

---

Would you like further details or explanations? Here are 5 related questions for deeper exploration:

1. What happens if we introduce friction between the block and the table?
2. How would the system behave if the hanging block were heavier than the block on the table?
3. Can this system reach equilibrium, and if so, under what conditions?
4. How does changing the mass of one of the blocks affect the tension in the rope?
5. How would the acceleration change if the pulley had significant mass?

**Tip:** Always double-check whether external forces like friction or air resistance are present, as they can significantly change the results.

Terdapat 2 buah balok yang tersusun seperti pada di gambar. Berat balok yang terletak di meja adalah 422 N, sedangkan yang tergantung adalah 185 N. Jika seluruh gaya gesek dapat diabaikan, tentukan (a) Percepatan dari kedua balok (b) tegangan tali sistem berikut.

Learn how to calculate the acceleration and tension in a two-block pulley system with weights of 422 N and 185 N. The solution involves using Newton's Second Law and key kinematics concepts. Suitable for students in Grades 10-12.

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