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Answer:
A. Coefficient of kinetic friction, μ = 0.34
B. The box moves a distance of 9.64 m before coming to a stop
C. The heavier box will stop after 2.1 seconds
Explanation:
a. The coefficient of kinetic or sliding friction is given as: μ = F/R
where applied force, F = m × (∆v)/t
∆v = v - u
∆v = 0 - 8 m/s = -8 m/s; t = 2.4 s
R = normal reaction = m×g
where g = 9.8 m/s²
Substituting in the kinetic friction formula; μ = m∆v/t ÷ 1/m×g
μ = ∆v/g×t
μ = 8 / 9.8 × 2.4
μ = 0.34
b. Using the equation v² = u² + 2as to calculate the distance travelled by the box
where v = 0 m/s; u = 8.0 m/s; a = ? s = ?
From F = ma = μR
a = μR/m = (μ × m × g)/m
a = μg
a = 0.34 × 9.8
a = 3.32 m/s²
This is negative acceleration or deceleration
Substituting in the equation of motion
8² + 2 × -3.32 × s = 0
-6.64s = -64
s = 9.64 m
Therefore, the box moves a distance of 9.64 m before coming to a stop
c. The coefficient of friction is independent of mass.
Using the formula in (a): μ = ∆v/g×t
t = ∆v/μg
t = 7/0.34 × 9.8
t = 2.10 s
Therefore, the heavier box will stop after 2.1 seconds
The coefficient of kinetic friction of the box is 0.34.
The distance traveled by the box is 9.6 m.
The time taken for the heavier box to stop is 2.1 s.
The coefficient of kinetic friction of the box is calculated as follows;
[tex]\mu mg = ma\\\\\mu g = a\\\\\mu g = \frac{v}{t} \\\\\mu = \frac{v}{gt} \\\\\mu = \frac{8}{9.8 \times 2.4} \\\\\mu = 0.34[/tex]
The distance traveled by the box is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\2as = -u^2\\\\s = \frac{-u^2}{2a} \\\\s = \frac{-u^2}{2\mu g} \\\\s = \frac{-(8)^2}{2\times 0.34 \times 9.8} \\\\s = -9.6 \ m\\\\|s| = 9.6 \ m[/tex]
The time taken for the heavier box to stop is calculated as;
[tex]\mu = \frac{v}{gt} \\\\t = \frac{v}{\mu g} \\\\t = \frac{7}{0.34 \times 9.8}\\\\ t = 2.1 \ s[/tex]
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