Answer:
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,
[tex]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[/tex]
Since M cannot be greater than M₁ for M to move over M₁, we take the smaller number.
So, M = 3.66 kg
