Question45. (II) The block shown in Fig. $4-45$ has mass $m=7.0 \mathrm{~kg}$ and lies on a fixed smooth frictionless plane tilted at an angle $\theta=22.0^{\circ}$ to the horizontal. (a) Determine the acceleration of the block as it slides down the plane. (b) If the block starts from rest 12.0 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline?
FIGURE 4-45 Block on inclined plane. Problems 45 and 46.

Studdy Solution

STEP 1

1. The block has a mass $m = 7.0 \, \text{kg}$ .
2. The plane is smooth and frictionless.
3. The plane is tilted at an angle $\theta = 22.0^\circ$ to the horizontal.
4. The block starts from rest.
5. The distance up the plane from the base is $12.0 \, \text{m}$ .

STEP 2

1. Determine the acceleration of the block as it slides down the plane.
2. Calculate the block's speed at the bottom of the incline.

STEP 3

Determine the acceleration of the block as it slides down the plane.
The force causing the block to accelerate down the plane is the component of gravitational force parallel to the plane. This force is given by $F_{\parallel} = m g \sin \theta$ , where $g = 9.81 \, \text{m/s}^2$ is the acceleration due to gravity.
The acceleration $a$ of the block is given by Newton's second law $F = ma$ . Therefore, $a = g \sin \theta$ .
Calculate $a$ : $a = 9.81 \, \text{m/s}^2 \times \sin(22.0^\circ)$

STEP 4

Calculate $\sin(22.0^\circ)$ and then $a$ .
$\sin(22.0^\circ) \approx 0.3746$
$a = 9.81 \, \text{m/s}^2 \times 0.3746 \approx 3.67 \, \text{m/s}^2$
The acceleration of the block is $3.67 \, \text{m/s}^2$ .

STEP 5

Calculate the block's speed at the bottom of the incline.
Use the kinematic equation for motion along a straight line: $v^2 = u^2 + 2as$
where: - $v$ is the final velocity, - $u = 0 \, \text{m/s}$ is the initial velocity (since the block starts from rest), - $a = 3.67 \, \text{m/s}^2$ is the acceleration, - $s = 12.0 \, \text{m}$ is the distance traveled down the plane.
Substitute the known values: $v^2 = 0 + 2 \times 3.67 \, \text{m/s}^2 \times 12.0 \, \text{m}$

STEP 6

Calculate $v^2$ and then $v$ .
$v^2 = 2 \times 3.67 \times 12.0 = 88.08$
$v = \sqrt{88.08} \approx 9.38 \, \text{m/s}$
The block's speed when it reaches the bottom of the incline is approximately $9.38 \, \text{m/s}$ .

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms