Question

A block with mass m1 = 2.30 kg rests on a frictionless table. It is connected with a light string over a pulley to a hanging block of mass m2 = 4.30 kg. The pulley is a uniform disk with a radius of 3.30 cm and a mass of 0.500 kg. 1. Calculate the acceleration of each block. (Express your answer to three significant figures.) 2. Calculate the tension in the x-direction (horizontal part of string). (Express your answer to three significant figures.) 3. Calculate the tension in the y-direction (vertical part of string). (Express your answer to three significant figures.) 4. How long does it take the blocks to move a distance of 2.45 m? (Express your answer to three significant figures.) 5) What is the angular speed of the pulley at this time? (Express your answer to three significant figures.)

Answer 1

1. Acceleration of each block:
`\( 2.98 \, \text{m/s}^2 \)`

2. Tension in the x-direction:
`\( 29.39 \, \text{N} \)`

3. Tension in the y-direction:
`\( 42.21 \, \text{N} \)`

4. Time to move 2.45 m:
`\( 1.28 \, \text{s} \)`

5. Angular speed of the pulley:
`\( 57.88 \, \text{rad/s} \)`

1. Calculate the acceleration of each block:

The acceleration can be found using Newton's second law ( F = ma ) and considering the tension in the string.

The system's equations:

For mass
`\( m_1 \)`:

`\[ T - m_1g = m_1a \]`

For mass
`\( m_2 \)`:

`\[ m_2g - T = m_2a \]`

Solving these equations simultaneously gives the acceleration:

`\[ a = (m_2g - m_1g)/(m_1 + m_2) \]`

Let's calculate it:

Given:
`\( m_1 = 2.30 \, \text{kg} \), \( m_2 = 4.30 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \)`

`\[ a = ((4.30 * 9.81) - (2.30 * 9.81))/(2.30 + 4.30) \]`

`\[ a \approx (42.213 - 22.533)/(6.60) \]`

`\[ a \approx (19.68)/(6.60) \]`

`\[ a \approx 2.98 \, \text{m/s}^2 \]`

2. Calculate the tension in the x-direction:

The tension in the horizontal direction is the same for both blocks since it's a light string.

`\[ T = m_1a + m_1g \]`

`\[ T = 2.30 * 2.98 + 2.30 * 9.81 \]`

`\[ T \approx 6.85 + 22.54 \]`

`\[ T \approx 29.39 \, \text{N} \]`

3. Calculate the tension in the y-direction:

The tension in the vertical direction supports the weight of
`\( m_2 \)`.

`\[ T = m_2g \]`

`\[ T = 4.30 * 9.81 \]`

`\[ T \approx 42.21 \, \text{N} \]`

4. How long does it take the blocks to move a distance of 2.45 m:

We'll use the equation
`\( s = ut + (1)/(2)at^2 \) where \( u = 0 \)` as they start from rest.

`\[ s = (1)/(2)at^2 \]`

`\[ t = \sqrt{(2s)/(a)} \]`

Given:
`\( s = 2.45 \, \text{m} \), \( a = 2.98 \, \text{m/s}^2 \)`

`\[ t = \sqrt{(2 * 2.45)/(2.98)} \]`

`\[ t \approx √(1.65) \]`

`\[ t \approx 1.28 \, \text{s} \]`

5. What is the angular speed of the pulley:

The linear speed of the blocks is the same as the tangential speed of the pulley.

`\[ v = r \omega \]`

`\[ \omega = (v)/(r) \]`

Given:
`\( r = 0.033 \, \text{m} \), \( s = 2.45 \, \text{m} \), \( t = 1.28 \, \text{s} \)`

`\[ v = (s)/(t) \]`

`\[ v = (2.45)/(1.28) \]`

`\[ v \approx 1.91 \, \text{m/s} \]`

`\[ \omega = (1.91)/(0.033) \]`

`\[ \omega \approx 57.88 \, \text{rad/s} \]`