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Inessa05 [86]
3 years ago
8

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 21 kg and the larger bottom crate has a m

ass of m2 = 90 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.8 and the coefficient of kinetic friction between the two crates is μk = 0.64. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem). 1)The rope is pulled with a tension T = 261 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate? m
Physics
1 answer:
Andrews [41]3 years ago
8 0

Answer:

2.35 m/s²

Explanation:

Given that

Mass of the smaller crate, m₁ = 21 kg

Mass of the larger crate, m₂ = 90 kg

Tensión of the rope, T = 261 N

We know that the sum of all forces for the two objects with a force of friction F and a tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F, where m and a are the masses and accelerations respectively.

1) no sliding can also mean that:

a₁ = a₂ = a

This makes us merge the two equations written above together as:

m₂a = T - m₁a

If we then solve for a, we would have something like this

a = T / (m₁+m₂)

a = 261 / (21 + 90)

a = 261 / 111

a = 2.35 m/s²

Therefore, the needed acceleration of the small crate is 2.35 m/s²

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Answer:

10

Explanation:

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2 years ago
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Answer:

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q = +2e

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3 years ago
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Shtirlitz [24]

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d = 90 ft

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d = h + 2(\frac{4}{5}h) + 2(\frac{4}{5})^2h + 2(\frac{4}{5})^3h........

here we know that

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2 years ago
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