Question

in the figure shown, the coefficient of kinetic friction between the block and the incline is 0.40. what is the magnitude of the acceleration of the suspended block as it falls? disregard any pulley mass or friction in the pulley.

Answer 1

The magnitude of the acceleration of the suspended block as it falls is 3.4 m/
`s^(2)` (approximately).

The magnitude of the acceleration of the suspended block as it falls can be determined using the formula:

acceleration = g * (sin(θ) -
`u_(k)` * cos(θ))

Where,

g is the acceleration due to gravity

θ is the angle of the incline

`u_(k)` is the coefficient of kinetic friction.

In this case, since there is no pulley mass or friction in the pulley, the acceleration will be solely determined by the angle of the incline and the coefficient of kinetic friction.

By substituting the given values, the magnitude of the acceleration of the suspended block can be calculated as:

acceleration = g * (sin(θ) -
`u_(k)` * cos(θ))

= 9.8 * (sin (21.7) - 0.4 * cos (21.7))

= 3.4 m/
`s^(2)` (approximately)

Answer 2

The magnitude of the acceleration of the suspended block as it falls is approximately
`3.4 \text m \text / \text s^2`

How to determine the magnitude of the acceleration?

The net force causing the block's acceleration is expressed using Newton's second law as:

`\[ ma = mg \sin(\theta) - \mu_k mg \cos(\theta) \]`

Where:

`\(m\) represents the mass of the block (64.4 kg)`.

`\(g\) stands for the acceleration due to gravity (9.81 m/s^2)`.

`\(\theta\) denotes the angle of the incline (21.7\textdegree)`.

`\(\mu_k\) indicates the coefficient of kinetic friction (0.40)`.

When applying the given values:

`\[ a = (64.4 \, \text{kg})(9.81 \, \text{m/s}^2) \sin(21.7^\circ) - (0.40)(64.4 \, \text{kg})(9.81 \, \text{m/s}^2) \cos(21.7^\circ) \]`

Solving for a:

`\[ a \approx 3.4 \, \text{m/s}^2`]

Therefore, the magnitude of the acceleration of the suspended block as it descends the incline is approximately
`\(3.4 \, \text{m/s}^2\)`.