Respuesta :
Answer:
0.0367
Explanation:
The loss in kinetic energy results into work done by friction.
Since kinetic energy is given by
KE=0.5mv^{2}
Work done by friction is given as
W= umgd
Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.
Making u the subject of the formula then we deduce that
[tex]u=\frac {v^{2}}{2gd}[/tex]
Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then
[tex]u=\frac {1.2^{2}}{2*9.81*2}=0.0366972477064\approx 0.0367[/tex]
Therefore, the coefficient of kinetic friction is approximately 0.0367
If a baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s, the coefficient of kinetic friction = 0.367
The mass of the suitcase, m = 15 kg
The speed of the compartment, v = 1.2 m/s
The distance moved by the baggage, d = 2.0 m
Note that the workdone in moving the suitcase is given by the formula:
W = μmgd.....................(1)
The kinetic energy is given by the formula:
[tex]KE = \frac{1}{2}mv^2[/tex].....................(2)
Note that the workdone = Kinetic Energy
[tex]\mu mgd = \frac{1}{2}mv^2\\\\\mu gd = 0.5v^2[/tex]
Substitute d = 0.2m, v = 1.2 m/s and g = 9.8 m/s² into the equation above
[tex]\mu \times 9.8 \times 2 = 0.5 \times 1.2^2\\\\19.6 \mu = 7.2\\\\\mu = \frac{7.2}{19.6} \\\\\mu = 0.367[/tex]
The coefficient of kinetic friction = 0.367
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