Question

A block with mass m1 = 7.5 kg rests on the surface of a horizontal table which has a coefficient of kinetic friction of μk = 0.61. A second block with a mass m2 = 11.4 kg is connected to the first by an ideal string passing over an ideal pulley such that the second block is suspended vertically. The second block is released from rest, and motion occurs. Block 1 accelerates along the tabletop, in the horizontal direction, while block 2 moves vertically. With the coordinate system provided in the drawing, we may write a-1 = a1 i and a-2 = a2 y. Write an expression that relates the vertical component of the acceleration of block 2 to the horizontal component of the acceleration of block 1.

Answer 1

The vertical acceleration of block 2 can be related to the horizontal acceleration of block 1 using Newton's second law and the concept of tension.

Step-by-step explanation:

The vertical component of the acceleration of block 2 can be related to the horizontal component of the acceleration of block 1 using Newton's second law and the concept of tension. Since block 2 is suspended vertically and connected to block 1 by a string passing over a pulley, the tension in the string will be the same for both blocks.

Let's assume the vertical acceleration of block 2 is a2 and the horizontal acceleration of block 1 is a1. Then we can write:

a2 = a1 * (m1/m2)

where m1 is the mass of block 1 and m2 is the mass of block 2. This equation shows that the vertical component of acceleration is directly proportional to the horizontal component of acceleration.

Answer 2

This expression indicates that the acceleration of both blocks is influenced by the tension in the string and the gravitational acceleration
`\( g \) (9.81 m/s²)` , adjusted by the mass of block 1 and the coefficient of kinetic friction.

To relate the vertical component of the acceleration of block 2 to the horizontal component of the acceleration of block 1, we can use Newton's second law and the constraints of the pulley system.

Since the string is ideal (massless and inextensible), the magnitude of the acceleration of both blocks must be the same. Therefore, the vertical acceleration of block 2
`(\( a_2 \))` is equal in magnitude to the horizontal acceleration of block 1
`(\( a_1 \)),` but they are in different directions.

Step-by-step approach:

1. Write the force equation for block 1 (horizontal direction):

The only force acting on block 1 in the horizontal direction is the tension in the string
`(\( T \))` minus the frictional force. The frictional force is given by
`\( F_(friction)` =
`\mu_k \cdot F_(normal) \)`, where
`\( \mu_k \)` is the coefficient of kinetic friction and
`\( F_(normal) \)` is the normal force, which equals the gravitational force on block 1,
`\( m_1 \cdot g \).`

So, for block 1:

`\[ T - \mu_k \cdot m_1 \cdot g = m_1 \cdot a_1 \]`

2. Write the force equation for block 2 (vertical direction):

For block 2, the forces are the gravitational force
`\( F_(normal) \)` and the tension in the string. The net force is
`\( m_2 \cdot g - T \).`

So, for block 2:

`\[ m_2 \cdot g - T = m_2 \cdot a_2 \]`

3. Relate the accelerations:

Since the string is ideal,
`\[ m_2 \cdot g - T = m_2 \cdot a_2 \]` in magnitude. We can use this relation to eliminate
`\( T \)` from the equations and solve for
`\( a_1 \) or \( a_2 \).`

Let's perform these steps to write an expression that relates
`\( a_2 \) to \( a_1 \).`

The expression relating the vertical component of the acceleration of block 2
`(\( a_2 \))` to the horizontal component of the acceleration of block
`1 (\( a_1 \))` in the system described is:

`\[ a_2 = a_1 = 9.81 - 0.0877 * T \]`

where
`\( T \)` is the tension in the string.

This expression indicates that the acceleration of both blocks is influenced by the tension in the string and the gravitational acceleration
`\( g \) (9.81 m/s²)` , adjusted by the mass of block 1 and the coefficient of kinetic friction.

The reduction in acceleration due to the tension and friction on block 1 is captured in the term
`\( 0.0877 * T \). Since \( a_1 \) and \( a_2 \)` are equal in magnitude (but in different directions) for an ideal string and pulley system, this expression effectively relates the two accelerations.