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ch4aika [34]
3 years ago
10

An inclined plane of effective length 4.5m is used to raise a load of 500N through a height of 1.5m .If the list is raise by for

ce of 200N
calculate frictional force between load and the plane​
Physics
1 answer:
IrinaK [193]3 years ago
8 0

Answer:

1850 N

Explanation:

The formula for friction force between the load and plane is given as ;

F= μ*N

N = mg cos θ

To find θ, which is the angle the inclined plane makes with the ground at the height of 1.5 m

Sin θ = 1.5/4.5

Sin θ = 0.3333

Sin⁻{0.3333} = 19.50°

θ = 19.50°

Finding N , where m= 500 N , and g= 9.81

N = mg cos θ

N= 500 * 9.81 * cos 19.50°

N= 4624 N

Coefficient of kinetic friction is calculated as;

μ=F/W

μ = 200/500 = 0.4

The magnitude of kinetic friction is given as;

Fk= μ * N

Fk = 0.4 * 4624

Fk= 1850 N

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Answer:

Explanation:

average velocity is 41.6/1.89 = 22.0 m/s

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In process numbers have been rounded to 3 s.d. for reporting purposes.

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Read 2 more answers
A uniform wooden plank with a mass of 75kg and length of 5m is placed on top of a brick wall so that 1.5m of plank extends beyon
jek_recluse [69]

Answer:

x₂ = 1.33 m

Explanation:

For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall

            Στ = 0

            W₁ x₁ - W₂ x₂ = 0

where W₁ is the weight of the woman, W₂ the weight of the table.

Let's find the distances.

Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.

           x₁ = 2.5 -1.5 = 1 m

The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative

            x₂ = \frac{M_1g  }{m_2 g} \ x_1

           

let's calculate

           x₂ = \frac{100}{75}  \ 1

           x₂ = 1.33 m

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