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3 years ago

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A 5.0-kg box rests on a horizontal surface. The coefficient of kinetic friction between the box and surface is ????K=0.50. A hor

izontal force pulls the box at constant velocity for 10 cm. Find the work done by (a) the applied horizontal force, (b) the frictional force, and (c) the net force.

1 answer:

blagie [28]3 years ago

3 0

Answer:

a)W= 1 J

b)W= 1 J

c)W= 0 J

Explanation:

Given that

m = 5 kg

Coefficient of kinetic friction,K=0.5

d= 10 cm

We know that if velocity is constant it means that the acceleration of the system is zero.Or we can say that all forces are balance in the system.

We know that work is the dot product of force and displacement

W= F.d

The friction force on the box

Fr= K m g

Fr= 0.5 x 5 x 10 = 25 N ( take g =10 m/s²)

Fr=25 N

a)

Acceleration a= 0

So horizontal force F = Fr

W = Fr.d

W=10 x 0.1 J

W= 1 J

b)

The work done by friction force

W= 1 J

c)

The net force on the system is zero because acceleration is zero.

F= 0

So

W= 0 J

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The described situation is known as Refraction.

Refraction is a phenomenon in which a wave (the light in this case) bends or changes it direction when passing through a medium with a refractive index different from the other medium.

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$n_{2}$ is the second medium refractive index

$\theta_{1}$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

Knowing this, let's begin with the answers:

<h2><u>1) Indexes of refraction for red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=38.1\º$ :

$(1)sin(57.0\º)=n_{2}sin(38.1\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(38.1\º)}$

$n_{2}=1.359$ (3)>>>Index of Refraction for red light

<h2>1b) Violet light</h2>

Again, using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=36.7\º$ :

$(1)sin(57.0\º)=n_{2}sin(36.7\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(36.7\º)}$

$n_{2}=1.403$ (4) >>>Index of Refraction for violet light

<h2><u>2) Speeds of red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Here we are going to use equation (1):

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Substituting (3) in this equation:

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<h2>1a) Violet light</h2>

Using again equation (1):

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Substituting (4) in this equation:

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