Physics, 04.07.2020 22:01 TACOSDAILY

A block of mass M rests on a block of mass M1 which is on a tabletop. A light string passes over a frictionless peg and connects the blocks. The coefficient of kinetic friction between the blocks and between M1 and the tabletop is the same. A force F pulls the upper block to the left and the lower block to the right. The blocks are moving at a constant speed. Required:Determine the mass of the upper block. (Express your answer to three significant figures.)

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Answer from: animegirl02

M = 3.66 kg

Explanation:

Here is the complete question

A block of mass M rests on a block of mass M1 = 5.00 kg which is on a tabletop. A light string passes over a frictionless peg and connects the blocks. The coefficient of kinetic friction μ

k at both surfaces equals 0.330. A force of F = 56.0 N pulls the upper block to the left and the lower block to the right. The blocks are moving at a constant speed. Determine the mass of the upper block. (Express your answer to three significant figures.)

Solution

The forces on mass M are

F - μMg = Ma (1)

The forces on mass M₁ are

F - μ(M + M₁)g = M₁a (2) (since both weights act downwards on M)

From (1) a = (F - μMg)/M

Substituting a into (2), we have

F - μ(M + M₁)g = M₁((F - μMg)/M)

Cross-multiplying M we have

MF - μ(M + M₁)Mg = M₁F - μMM₁g

Expanding the bracket, we have

MF - μM²g + μM₁Mg = M₁F - μMM₁g

We now collect like terms

MF - μM²g + μM₁Mg + μMM₁g = M₁F

MF - μM²g + 2μM₁Mg - M₁F = 0

- μM²g + 2μM₁Mg + MF - M₁F = 0

Dividing through by -1, we have

- μM²g + (2μM₁g + F)M - M₁F = 0

μM²g - (2μM₁g + F)M + M₁F = 0

M² - (2M₁ + F/μg)M + M₁F/μg = 0

We now have a quadratic equation in M. We now substitute the values of the variables int o the quadratic equation to get

M² - (2(5 kg) + 56 N/(0.33 × 9.8 m/s²))M + (5 kg × 56 N)/(0.33 × 9.8 m/s²) = 0

M² - (10 kg) + 56 N/3.234 m/s²)M + (5 kg × 56 N)/(3.234 m/s²) = 0

M² - (10 kg + 17.32 kg) M + 86.58 kg = 0

M² - 27.32 kg M + 86.58 kg = 0

Using the quadratic formula

with a = 1, b = -27.32 and c = 86.58,

$M = \frac{-(-27.32) +/- \sqrt{(-27.32)^{2} - 4 X 1 X 86.58} }{2 X 1} \\= \frac{27.32 +/- \sqrt{746.38 - 346.32} }{2}\\= \frac{27.32 +/- \sqrt{400.06} }{2}\\= \frac{27.32 +/- 20.001 }{2}\\= \frac{27.32 + 20.001 }{2} or \frac{27.32 - 20.001 }{2}\\= \frac{47.32 }{2} or \frac{7.32 }{2}\\\\=23.66 or 3.66$

Since M cannot be greater than M₁ for M to move over M₁, we take the smaller number.

So, M = 3.66 kg

Answer from: Quest

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample, $v = 1.1 cm^{3}$

so, density of rock sample will be

$density = \frac{mass}{volume}$

$\rho =\frac{2.7}{1.1}g/cm^{3}\\\\\rho =2.45 g/cm^{3}$

since, the density of rock sample is $2.45 g/cm^{3}$ . hence, the rock sample is a sedimentary rock, whose densities varies between $2g/cm^{3} - 2.6 g/cm^{3}$ .

Answer from: Quest

c) up

hope this

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