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Nataly_w [17]
2 years ago
13

you slide abox of books at constant speed up a30 degree ramp, applying a force of 200 newtons directed up the slope. The coeffic

ient of sliding friction is 0.18. How much work have you done when the box has risen 1meter vertically
Physics
1 answer:
Tomtit [17]2 years ago
5 0

The work done is 400 J

Explanation:

The work done by you in pushing the box along the slope is given by

W=Fd

where

F is the magnitude of the force applied

d is the distance covered by the box along the slope

Here we have the following:

F = 200 N is the magnitude of the force applied

d=\frac{h}{sin \theta} is the distance covered, where

h = 1 m is the vertical rise

\theta=30^{\circ} is the slope of the plane

Substituting and solving, we find

W=F\frac{h}{sin \theta}=\frac{(200)(1)}{sin 30^{\circ}}=400 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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To solve the problem it is necessary to identify the equation in the manner given above.

This equation corresponds to the displacement of a body under the principle of simple harmonic movement.

Where,

\xi = Acos(\omega t +\phi)

PART A) Our equation corresponds to

y = -5cos(4\pi t)

Therefore the value of omega is equivalent to that of

\omega = 4\pi

From the definition we know that the period as a function of angular velocity is equivalent to

T = \frac{2\pi}{\omega}

T = \frac{2\pi}{4\pi}

T = \frac{1}{2}

This same point is the equivalent of the maximum point of the speed that the body can reach, since the internal expression of the cos\thetaIs equivalent to . So the maximum speed that the body can reach is,

y = -5cos(4\pi t)

y = -5cos(4\pi (1/2))

y = -5*(-1)

y = 5

Therefore the maximum felocity will be 5ft / s

PART B) The period of graph is the time taken to reach from one maximum point to next point maximum point, then

t = \frac{T}{2} = \frac{1}{2}*\frac{1}{2}

t = \frac{1}{4}s

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3 years ago
What type of fingerprints is invisible to the naked eye?
luda_lava [24]
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P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism ar
creativ13 [48]

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

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