Question

A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.

A)Determine the acceleration of the crate as it slides down the plane.
Express your answer to two significant figures and include the appropriate units.

B)If the crate starts from rest 8.70 m up along the plane from its base, what will be the crate's speed when it reaches the bottom of the incline?
Express your answer to two significant figures and include the appropriate units.

Answer 1

The acceleration of the crate as it slides down the plane is approximately 4.86 m/s². The speed of the crate when it reaches the bottom of the incline is approximately 4.41 m/s.

Step-by-step explanation:

To determine the acceleration of the crate as it slides down the plane, we need to consider the forces acting on the crate. The force of gravity can be divided into two components: one parallel to the inclined plane and one perpendicular to the plane. The component parallel to the plane is given by F_parallel = m * g * sin(theta), where m is the mass of the crate, g is the acceleration due to gravity, and theta is the angle of the plane. The frictional force acting on the crate is given by F_friction = mu_k * m * g * cos(theta), where mu_k is the coefficient of kinetic friction. The net force, F_net, acting on the crate parallel to the plane is given by F_net = F_parallel - F_friction. The acceleration, a, of the crate is then given by a = F_net / m.

Using the given values (m = 20.0 kg, theta = 30.0°, mu_k = 0.0300, g = 9.8 m/s²), we can calculate the acceleration:

F_parallel = (20.0 kg) * (9.8 m/s²) * sin(30.0°) ≈ 98 N

F_friction = (0.0300) * (20.0 kg) * (9.8 m/s²) * cos(30.0°) ≈ 0.831 N

F_net = 98 N - 0.831 N ≈ 97.2 N

a = (97.2 N) / (20.0 kg) ≈ 4.86 m/s²

Therefore, the acceleration of the crate as it slides down the plane is approximately 4.86 m/s².

To determine the speed of the crate when it reaches the bottom of the incline, we can use the following kinematic equation: v² = u² + 2as, where v is the final velocity of the crate, u is the initial velocity (0 m/s since the crate starts from rest), a is the acceleration, and s is the distance traveled (2 m). Plugging in the values, we have v² = 0² + 2 * (4.86 m/s²) * (2 m) = 19.44 m²/s². Taking the square root of both sides, we find v ≈ 4.41 m/s. Therefore, the speed of the crate when it reaches the bottom of the incline is approximately 4.41 m/s.

Answer 2

A)
`2.03 m/s^2`

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

`mg sin \theta - \mu_k R = ma` (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

`\theta=22.5^(\circ)`

`\mu_k = 0.19` is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

`R-mg cos \theta =0` (2)

From (2) we find

`R=mg cos \theta`

And substituting into (1)

`mg sin \theta - \mu_k mg cos \theta = ma`

Solving for a,

`a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2`

B) 5.94 m/s

We can solve this part by using the suvat equation

`v^2-u^2=2as`

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

`a=2.03 m/s^2`

s = 8.70 m

Solving for v,

`v=√(u^2+2as)=√(0+2(2.03)(8.70))=5.94 m/s`