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Pie
3 years ago
5

A 2.4 kg box has an initial velocity of 3.6 m/s upward along a plane inclined at 27◦ to the horizontal. The coefficient of kinet

ic friction between the box and the plane is 0.12. The acceleration of gravity is 9.81 m/s 2 . How far up the incline does the box travel? Answer in units of m.
Physics
1 answer:
Vika [28.1K]3 years ago
7 0

Answer:

d= 1.18 m

Explanation:

In abscense of  friction, total mechanical energy must be constant, i.e.,

ΔK + ΔU = 0

As we are told that there exists a kinetic friction between the box and the plane, we need to take into account the work done by the friction force in the equation, as follows:

ΔK + ΔU = Wnc (1)

If we take as our zero gravitational potential energy reference, the height at which the box is sent upward, we can write the following equations for the different terms in (1):

ΔK = Kf- K₀ = 0 - 1/2*m*v₀² = -1/2*2.4kg* (3.6)²(m/s)² = -15.6 J

ΔU = Uf - U₀ = m*g*h = *m*g*d*sin θ = 2.4 kg*9.81 m/s²*d*0.454 = 10.7*d J

Wnc = Ff. d* cos (180º) = μk*N*d*cos(180º) (2)

The friction force always opposes to the displacement, so the angle between force and displacement is 180º.

The normal force, as is always perpendicular to the surface, takes the value needed to equilibrate the component of the weight perpendicular to the incline, as follows:

N = m*g*cosθ =  2.4 kg*9.81 m/s²*cos 27º = 21 N

Replacing in (2):

Wnc = 0.12*21*cos (180º) = -2.52*d J

Replacing in (1):

-15.6 J + 10.7*d J = -2.52*d J

Solving for d:

d = 1.18 m

 

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Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector  \hat{i}.
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The airplane flies at 340 km/h at 12° east of north. Its velocity vector is
\vec{v}_{1} = 340(sin(15^{o})\hat{i} + cos(15^{o})\hat{j} ) = 88\hat{i} + 328.4\hat{j}

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})

The plane's actual velocity is the vector sum of the two velocities. It is
\vec{v}=\vec{v}_{1}+\vec{v}_{2} = 121.1615\hat{i}+306.0473\hat{j}

The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h

The angle that the velocity makes north of east is
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1. Una bola de 0.510 kg de masa se mueve al este (dirección +x) con una rapidez de 4.80 m/s y choca frontalmente con una bola de
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Responder:

3,37 m / s, + ve x - dirección

Explicación:

Utilizando la ley de conservación de la cantidad de movimiento expresada por la fórmula;

m1u1 + m2u2 = (m1 + m2) v

m1 y m2 son las masas de los objetos

u1 y u2 son sus velocidades iniciales

v es su velocidad común

Dado

m1 = 0,519 kg

u1 = 4,80 m / s

m2 = 0,220 kg

u2 = 0 m / s (cuerpo en reposo)

Necesario

Velocidad común v

Sustituir en la fórmula los valores dados;

0,519 (4,8) + 0,22 (0) = (0,519 + 0,220) v

2,4912 + 0 = 0,739 v

2,4912 = 0,739v

Dividir ambos lados por 0,739

2,4912 / 0,739 = 0,739 / 0,739

<em>v = 3,37 m / s </em>

<em>Por lo tanto, la rapidez de ambas bolas después de la colisión es de 3.37 m.s hacia la dirección x positiva, ya que m1> m2 y la velocidad común es positiva.</em>

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The initial position of the object was found to be 134.09 m.

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As displacement is the measure of difference between the final and initial points. In other words, we can say that displacement can be termed as the change in the position of the object irrespective of the path followed by the object to change the path. So

Displacement = Final position - Initial position.

As the final position is stated as -55.25 meters and the displacement is also stated as -189.34 meters. So the initial position will be

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Thus, the initial position for the object having a displacement of -189.34 m is determined as 134.09 m.

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