Math / Algebra

Question43. Two blocks are connected by a massless rope as shown below. The mass of the block on the table is 4.0 kg and the hanging mass is 1.0 kg . The table and the pulley are frictionless. (a) Find the acceleration of the system. (b) Find the tension in the rope. (c) Find the speed with which the hanging mass hits the floor if it starts from rest and is initially located 1.0 m from the floor.

Studdy Solution

STEP 1

1. The system consists of two blocks connected by a massless rope.
2. The block on the table has a mass of 4.0 kg.
3. The hanging block has a mass of 1.0 kg.
4. The table and pulley are frictionless.
5. We need to find the acceleration of the system, the tension in the rope, and the speed of the hanging mass when it hits the floor.

STEP 2

1. Analyze the forces acting on each block.
2. Write equations of motion for each block.
3. Solve the equations to find the acceleration of the system.
4. Use the acceleration to find the tension in the rope.
5. Use kinematic equations to find the speed of the hanging mass when it hits the floor.

STEP 3

Analyze the forces acting on each block.
- For the block on the table (mass $m_1 = 4.0 \, \text{kg}$ ), the only horizontal force is the tension $T$ in the rope. - For the hanging block (mass $m_2 = 1.0 \, \text{kg}$ ), the forces are tension $T$ upward and gravitational force $m_2g$ downward.

STEP 4

Write equations of motion for each block.
- For the block on the table: $T = m_1a$ - For the hanging block: $m_2g - T = m_2a$

STEP 5

Solve the equations to find the acceleration of the system.
Add the two equations to eliminate $T$ :
$m_2g - T + T = m_2a + m_1a$ $m_2g = (m_1 + m_2)a$
Solve for $a$ :
$a = \frac{m_2g}{m_1 + m_2}$
Substitute $m_1 = 4.0 \, \text{kg}$ , $m_2 = 1.0 \, \text{kg}$ , and $g = 9.8 \, \text{m/s}^2$ :
$a = \frac{1.0 \times 9.8}{4.0 + 1.0}$ $a = \frac{9.8}{5.0}$ $a = 1.96 \, \text{m/s}^2$

STEP 6

Use the acceleration to find the tension in the rope.
Substitute $a = 1.96 \, \text{m/s}^2$ into the equation for the block on the table:
$T = m_1a$ $T = 4.0 \times 1.96$ $T = 7.84 \, \text{N}$

STEP 7

Use kinematic equations to find the speed of the hanging mass when it hits the floor.
Use the equation $v^2 = u^2 + 2as$ , where $u = 0 \, \text{m/s}$ (initial speed), $a = 1.96 \, \text{m/s}^2$ , and $s = 1.0 \, \text{m}$ :
$v^2 = 0 + 2 \times 1.96 \times 1.0$ $v^2 = 3.92$ $v = \sqrt{3.92}$ $v \approx 1.98 \, \text{m/s}$
The acceleration of the system is $1.96 \, \text{m/s}^2$ , the tension in the rope is $7.84 \, \text{N}$ , and the speed of the hanging mass when it hits the floor is approximately $1.98 \, \text{m/s}$ .

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Studdy Solution

STEP 1

STEP 2

1. Analyze the forces acting on each block.
2. Write equations of motion for each block.
3. Solve the equations to find the acceleration of the system.
4. Use the acceleration to find the tension in the rope.
5. Use kinematic equations to find the speed of the hanging mass when it hits the floor.

STEP 3

Analyze the forces acting on each block.
- For the block on the table (mass $m_1 = 4.0 \, \text{kg}$ ), the only horizontal force is the tension $T$ in the rope. - For the hanging block (mass $m_2 = 1.0 \, \text{kg}$ ), the forces are tension $T$ upward and gravitational force $m_2g$ downward.

STEP 4

Write equations of motion for each block.
- For the block on the table: $T = m_1a$ - For the hanging block: $m_2g - T = m_2a$

STEP 5

STEP 6

Use the acceleration to find the tension in the rope.
Substitute $a = 1.96 \, \text{m/s}^2$ into the equation for the block on the table:
$T = m_1a$ $T = 4.0 \times 1.96$ $T = 7.84 \, \text{N}$

STEP 7

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Studdy Solution

STEP 1

STEP 2

1. Analyze the forces acting on each block.2. Write equations of motion for each block.3. Solve the equations to find the acceleration of the system.4. Use the acceleration to find the tension in the rope.5. Use kinematic equations to find the speed of the hanging mass when it hits the floor.

STEP 3

STEP 4

Write equations of motion for each block.- For the block on the table: T=m1a T = m_1a T=m1​a - For the hanging block: m2g−T=m2a m_2g - T = m_2a m2​g−T=m2​a

STEP 5

STEP 6

Use the acceleration to find the tension in the rope.Substitute a=1.96 m/s2 a = 1.96 \, \text{m/s}^2 a=1.96m/s2 into the equation for the block on the table:T=m1a T = m_1a T=m1​a T=4.0×1.96 T = 4.0 \times 1.96 T=4.0×1.96 T=7.84 N T = 7.84 \, \text{N} T=7.84N

STEP 7

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1. Analyze the forces acting on each block.
2. Write equations of motion for each block.
3. Solve the equations to find the acceleration of the system.
4. Use the acceleration to find the tension in the rope.
5. Use kinematic equations to find the speed of the hanging mass when it hits the floor.

Write equations of motion for each block.
- For the block on the table: $T = m_1a$ - For the hanging block: $m_2g - T = m_2a$

Use the acceleration to find the tension in the rope.
Substitute $a = 1.96 \, \text{m/s}^2$ into the equation for the block on the table:
$T = m_1a$ $T = 4.0 \times 1.96$ $T = 7.84 \, \text{N}$