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3 years ago

A 50gk boy on a rough horizontal ground the coefficeint of static friction is 0.68

Physics

1 answer:

Soloha48 [4]3 years ago

4 0

Given that,

Mass of a boy = 50 kg

The coefficient of static friction = 0.68

To find,

Let us assume we need to find the maximum static friction between the boy and the ground.

Solution,

The formula for the maximum static friction between the two objects is given by :

$F=\mu N$

Where

N is normal force

Substitute all the values,

$F=0.68\times 50\times 10\\\\F=340\ N$

Therefore, the maximum static friction between the boy and the ground is 340 N.

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3 years ago

A spring with a spring constant of 1200 N/m has a 55-g ball at its end. The energy of the system is 5.5J. What is the amplitude

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3 years ago

A wheel of radius 0.38 m rotates in a clockwise sense about a fixed axle with negligible friction at an initial angular speed of

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Answer:

ωf = 2.19 rad/s

Explanation:

Newton's second law:

F = ma has the equivalent for rotation:

τ = I * α Formula (1)

where:

τ : It is the torque applied to the body. (Nxm)

I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Data

F = 24 N : Tangential force

m : 14 kg : mass of the wheel

R = 0.38 m : radius of the wheel

Moment of inertia of the wheel

The moment of inertia of wheel is defined as follows:

I = m* R²

I = 14 kg*(0.38)²

I = 2.0216 kg*m²

Torque applied to the wheel

The Torque ( τ) applied to the wheel is defined as follows:

τ = F*d

Where:

F : Tangential force applied to the wheel

d : Perpendicular distance of the tangential force to the axis of rotation

τ = (24N)*(0.38 m)

τ = 9.12 N*m

Angular acceleration of the wheel (α )

We replace data in the formula (1):

τ = I * α

9.12 = (2.0216) * α

α= 9.12 / (2.0216)

α = 4.5 rad/s²

Kinematics of the wheel

We apply the equations of circular motion uniformly accelerated :

ωf = ω₀ + α*t Formula (2)

Where:

α : Angular acceleration (rad/s²)

ω₀ : Initial angular speed ( rad/s)

ωf : Final angular speed ( rad

t : time interval (s)

Data

α = 4.5 rad/s²

ω₀ =1.6 rad/s

t = 0.13 s

We replace data in the formula (2):

ωf = ω₀ + α*t

ωf = 1.6 + (4.5)*(0.13)

ωf = 2.19 rad/s

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3 years ago