Answer:
The mass of the upper block M = 4.12 kg
Explanation:
From the given information:
The upper block formula can be computed as :
[tex]F = T + F_{f1}[/tex]
[tex]F - T - F_{f1} =0[/tex]
where;
[tex]F_{f1} = \mu Mg[/tex]
[tex]F - T - \mu Mg=0 ---- (1)[/tex]
The lower block formula is as follows:
[tex]T - F_{f1} -F_{f2} = 0[/tex]
[tex]T - \mu Mg - \mu (M_1+M)g= 0[/tex]
[tex]T = \mu Mg - \mu (M_1+M)g --- (2)[/tex]
Equating equation (1) and (2) together, we have:
[tex]F - \mu Mg - \mu Mg - \mu (M_1 + M) g = 0[/tex]
[tex]56 - 3 Mg - \mu M_1g= 0[/tex]
[tex]M = \dfrac{56 - \mu M_1 g}{3 \mu g}[/tex]
[tex]M = \dfrac{56 - (0.330)(5)(9.8)}{3 (0.330) (9.8)}[/tex]
[tex]M = \dfrac{56 - 16.17}{9.702}[/tex]
[tex]M = \dfrac{39.83}{9.702}[/tex]
M = 4.12 kg
We are given;
Coefficient of kinetic friction for both surfaces; μ_k = 0.33
Force; F = 56 N
Mass of lower block; m1 = 5 kg
From the image of this block and string system, we can find the sum of forces to get;
F = T + F_f1
Where;
T is tension in string
F_f1 is frictional force = μ_k × mg
Thus;
F = T + μ_k*mg - - - (1)
Where m is mass of the upper block
For the lower block, we can sum forces to get;
T = μ_k*mg + μ_k(m1 + m)g - - - (2)
Put μ_k*mg + μ_k(m1 + m)g for T in eq 1;
F = μ_k*mg + μ_k(m1 + m)g + μ_k*mg
F = 2μ_k*mg + μ_k(m1 + m)g
56 = (μ_k*g) * (2m + m1 + m)
56 = (0.33 * 9.8) * (3m + 5)
3m + 5 = 56/(0.33 * 9.8)
3m + 5 = 17.316
m = (17.316 - 5)/3
m = 4.11 kg
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