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2 years ago

A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.

Physics

1 answer:

Ray Of Light [21]2 years ago

8 0

The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>

Mass of the box, m = 3.1 kg
Distance moved by the box, d = 2.5 m
Coefficient of friction, = 0.35
Inclination of the force, θ = 30⁰

Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

$W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)$

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

$W = (0) d \times cos(30)\\\\W = 0 \ J$

The work done by the force of gravity is calculated as follows;

$W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J$

The work done on the box by the normal force is calculated as follows;

$W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J$

Learn more about work done here: brainly.com/question/8119756

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3 years ago

A disk of radius R = 7.52 cm is centered at the origin and lies along the y – z plane. The disk has a surface charge density σ =

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Answer:

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Explanation:

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3 years ago

A playground merry-go-round has radius 2.10m and moment of inertia 2500kg*m^2 about a vertical axle through its center, and it t

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Answer

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Time interval t = 16 s

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Let α be the angular acceleration.

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F r = Iα

Angular acceleration α = F r/I

= $\dfrac{18\times 2.1}{2500}$

= 0.015 rad/s²

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Final angular speed ω₂ = ω₁ + αt

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= 0.24 rad/s

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3 years ago

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Hope it helps:)

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3 years ago

Read 2 more answers