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makvit [3.9K]
3 years ago
14

A crate which has a mass of m1 = 125kg is sitting on an icy surface. A rope is attached to the crate and held at angle theta 28.

0 above the horizontal. If the rope is pulled by a force p= 585 N and the crate resultantly accelerates at a rate of A =3.30m/s squared , Find the coefficient of sliding friction between the crate and the icy surface.
Physics
1 answer:
devlian [24]3 years ago
3 0

Answer:

μ = 0.109

Explanation:

Draw a free body diagram of the crate.  There are four forces:

Weight force mg pulling down.

Normal force N pushing up.

Applied force P pulling at θ above the horizontal.

Friction force Nμ pushing to the left.

Sum of the forces in the y direction:

∑F = ma

N + P sin θ − mg = 0

N = mg − P sin θ

Sum of the forces in the x direction:

∑F = ma

P cos θ − Nμ = ma

P cos θ − ma = Nμ

μ = (P cos θ − ma) / N

μ = (P cos θ − ma) / (mg − P sin θ)

Given:

P = 585 N

θ = 28.0°

m = 125 kg

a = 3.30 m/s²

μ = (585 cos 28.0° − 125 kg × 3.30 m/s²) / (125 kg × 9.8 m/s² − 585 sin 28.0°)

μ = 0.109

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allsm [11]

This question is incomplete, the complete question;

you make an interferometer using 50-50 beam splitter and two mirrors, one being a perfect mirror and one which does not reflect all light. The wavelength of the 9 mW incident laser is 400 nm.

Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity

Options;

a) 200 nm; 0.9 mW

b) 100 nm, 0.0059 mW

c) 200 nm; 0 mW

d) 100 nm; 0.9 mW

e) 200 nm; 0.0059 mW

Answer:

the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector  and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer

Explanation:

Given that the instrument here is an interferometer.

Maximum intensity is obtained when the two waves are exactly in phase.

that is the peaks (crusts and troughs) and nodes (zero value points) of the two waves will be at the exact same point when the wave falls on the detector.

The phase factor of this point is taken as ∅ = 0

Now, to get a minimum point, the phase difference between the two waves should be should be ∅ = π

This corresponds to a path difference between the two waves as half of the wavelength. λ/2

The light gets reflected from the mirror.

Hence, when we move the mirror by a length l, the extra/less path the light has to travel is 2l (light is going and coming back)

hence, to get a path difference of λ/2 the mirror should move half of this distance only

so, the mirror should move;

l = λ/4

here, wavelength is 400nm

the length moved by the mirror = 400/4 = 100 nm

The intensity is given by the equation;

l = l1 + l2 + 2√l1l2cos(∅)

where

l1 = 2.25 mW

l2 = 2.025 mW

∅ = π

so we substitute

l = 2.25 + 2.025 - 2√(2.25 × 2.025)

l = 4.275 - 4.26907

l = 0.0059

Therefore; the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector  and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer  

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