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Crank
3 years ago
10

A student attaches a rope to a 20.0 kg box of books. He pulls with a force of 90.0 N at an angle of 30 degrees with the horizont

al. To make his job easier, he oiled the bottom of the box, reducing the coefficient of friction to 0.250. What is the magnitude of the acceleration of the box?

Physics
1 answer:
kolbaska11 [484]3 years ago
8 0

Answer:

The magnitude of the acceleration of the box is 2.01 m/s².

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

We have the following horizontal forces:

Fr = friction force.

Fx = Horizontal component of the applied force, F.

And we have the following vertical forces:

Fy = vertical component of the applied force.

N = normal force exerted on the box.

W = weight of the box.

According to Newton´s second law:

∑F = m · a

Then, in the horizontal direction:

Fx - Fr = m · a

Where "m" is the mass of the box and "a" its acceleration.

Fx can be obtained by trigonometry (see figure):

Fx = F · cos 30°

Fx = 90.0 N · cos 30°

Fr is calculated as follows:

Fr = μ · N

Where μ is the coefficient of friction and N the normal force.

So, we have to find the magnitude of the normal force.

Using Newton´s second law in the vertical direction:

∑F = N + Fy - W = m · a

Notice that the box has no vertical acceleration, then:

N + Fy - W = 0

Solving for N:

N = W - Fy

The weight is calculated as follows:

W = m · g

Where g is the acceleration due to gravity:

W = 20.0 kg · 9.8 m/s² = 196 N

And the vertical component of the applied force can be obtained by trigonometry:

Fy = F · sin 30°

Fy = 90.0 N · sin 30°

The normal force will be:

N = W - Fy = 196 N - 90.0 N · sin 30°

N = 151 N

Now, we can calculate the friction force:

Fr = μ · N

Fr = 0.250 · 151 N

Fr = 37.8 N

And now, we can obtain the acceleration of the box:

Fx - Fr = m · a

(Fx - Fr) / m = a

(90.0 N · cos 30° - 37.8 N ) / 20.0 kg = a

a = 2.01 m/s²

The magnitude of the acceleration of the box is 2.01 m/s².

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2 years ago
A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____
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Magnification, m = -0.42

Explanation:

It is given that,

Height of diamond ring, h = 1.5 cm

Object distance, u = -20 cm

Radius of curvature of concave mirror, R = 30 cm

Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)

Using mirror's formula :

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}, f = focal length of the mirror

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{-20}

v = -8.57 cm

The magnification of a mirror is given by,

m=\dfrac{-v}{u}

m=\dfrac{-(-8.57)}{-20}

m = -0.42

So, the magnification of the concave mirror is 0.42. Thew negative sign shows that the image is inverted.

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3 years ago
In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo
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Answer:

Among those two medium, light would travel faster in the one with a reflection angle of 32^{\circ} (when light enters from the air.)

Explanation:

Let v_{1} denote the speed of light in the first medium. Let v_{\text{air}} denote the speed of light in the air. Assume that the light entered the boundary at an angle of \theta_{1} to the normal and exited with an angle of \theta_{\text{air}}. By Snell's Law, the sine of \theta_{1}\! and \theta_{\text{air}}\! would be proportional to the speed of light in the corresponding medium. In other words:

\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}.

When light enters a boundary at the critical angle \theta_{c}, total internal reflection would happen. It would appear as if the angle of refraction is now 90^{\circ}. (in this case, \theta_{\text{air}} = 90^{\circ}.)

Substitute this value into the Snell's Law equation:

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Rearrange to obtain an expression for the speed of light in the first medium:

v_{1} = v_{\text{air}} \cdot \sin(\theta_{1}).

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For 0 < \theta < 90^{\circ}, \sin(\theta) is monotonically increasing with respect to \theta. In other words, for \!\theta in that range, the value of \sin(\theta)\! increases as the value of \theta\! increases.

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3 years ago
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