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mina [271]
3 years ago
15

A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

What is the average friction force opposing its motion?
Physics
1 answer:
Mama L [17]3 years ago
4 0

Answer: f_{r} = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8m/s^{2}

f_{r} = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}

we slot in a into the equation below to get frictional force

mgsinФ - f_{r} = ma

3 * 9.8 * sin 37 - f_{r} = 3* 0.4

17.9633 - f_{r} =  1.2

f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

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A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Oiled Steel fry
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Answer:

N  = 6.67 N

Explanation:

The frictional or frictional force is a force that arises from the contact of two bodies and opposes movement.

The friction is due to imperfections and roughness, mainly microscopic, that exist on the surfaces of the bodies. Upon contact, these roughnesses engage with each other making movement difficult. To minimize the effect of friction, either the surfaces are polished or lubricated, since the oil fills the imperfections, preventing them from snagging.

As the frictional force depends on the materials and the force exerted on one another, its magnitude is obtained by the following expression:

f = μ*N    Formula (1)

where:  

f is the friction force  (N)

μ is the coefficient of friction

N is the normal force (N)

Data

f = 0.2 N : frictional force between the steel spatula and the Oiled Steel frying pan

μ = 0.03 :coefficient of kinetic friction between the two materials

Calculating of normal force

We replace data in the formula (1)

f = μ*N  

0.2  = 0.03*N  

N  = 0.2 / 0.03

N  = 6.67 N

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3 years ago
A football kicker kicks a ball with a mass of .42 kg. The average acceleration of the football was 14.8 m/s squared. How much fo
mrs_skeptik [129]

To find the force we use the formula,

F = ma , where m is mass and a acceleration

Using the formula,

F = ma

F = 0.42 x 14.8

F = 6.216 N / 6.22 N

Hope you liked the answer !

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2 years ago
The wavelength of red helium-neon laser light in air is 632.8 nm.(a) What is its frequency? Hz(b) What is its wavelength in glas
vekshin1
<h2>Answers:</h2>

The speed of a wave is given by:

v=f.\lambda  (1)

Where f is the frequency and  \lambda the wavelength.

In the case of light, its speed is:

c=f.\lambda (2)

On the other hand, the described situation is known as Refraction,   a phenomenon in which the light changes its direction when passing through a medium with a refractive index different from the other medium.  

In this context, the Refractive index n is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum (c=3(10)^{8}m/s) and the speed of light v in the second medium:

n=\frac{c}{v} (3)

In addition, as the light changes its direction, its wavelength changes as well:

n=\frac{\lambda_{air}}{\lambda_{glass}} (4)

Knowing this, let's begin with the answers:

<h2>a) Frequency</h2>

From equation (2) we can find f:

f=\frac{c}{\lambda}  (5)

Knowing that 1nm=(10)^{-9}m:

f=\frac{3(10)^{8}m/s}{632.8(10)^{-9}m}  

f=4.74(10)^{14}Hz}     (6)   >>>Frequency of the helium-neon laser light

<h2>b) Wavelength in glass</h2>

We already know the wavelength of the light in air \lambda_{air} and the index of refraction of the glass.

So, we only have to find the wavelength in glass \lambda_{glass} from equation (4):

\lambda_{glass}=\frac{\lambda_{air}}{n}

\lambda_{glass}=\frac{632.8(10)^{-9}m}{1.48}

\lambda_{glass}=427(10)^{-9}m=427nm   (7)   >>>Wavelength of the helium-neon laser light in glass

<h2>c) Speed in glass</h2>

From equation (3) we can find the speed vof this light in glass:

v=\frac{c}{n}

v=\frac{3(10)^{8}m/s}{1.48}

v=2.027(10)^{8}m/s   (8)  >>>Speed of the helium-neon laser light in glass

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