Physics, 14.10.2019 05:20 fseftrwer2522

Acrate 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.

Answers: 2

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We have that for the Question it can be said that the acceleration of the crate as it slides down the plane and the crate's speed when it reaches the bottom of the incline is

a=2.03m/s^2V_f=5.9m/s

From the question we are told

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.

Generally the equation for the Force is mathematically given as

$mgsin\theta-Fr=ma\\\\Where\\\\Fr=\mu_k N\\\\Fr=m*9.8*0.19*0.9.923\\\\Fr=1.72m\\\\Therefore\\\\mgsin\theta-Fr=ma\\\\a=mgsin22.5-1.72\\\\a=2.03\\\\$

Generally the equation for the Kinematics is mathematically given as

$V_f^2=v^2+2ad\\\\V_f=\sqrt{0^2+2*2.03*8.70}$

V_f=5.9m/s

Hence,the acceleration of the crate as it slides down the plane and the crate's speed when it reaches the bottom of the incline is

a=2.03m/s^2V_f=5.9m/s

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Answer from: keaton3410

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,

$v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s$

Answer from: Quest

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Answer from: Quest

He can play two games. 13/4=3.253.25*2=6.5

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

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