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 suitcase slides 2.0 m before stopping. Use work and energy to find the suitcase’s coefficient of kinetic friction on the floor.

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:

.....................(2)

Note that the workdone  =  Kinetic Energy

Substitute d = 0.2m, v = 1.2 m/s and g = 9.8 m/s² into the equation above

The coefficient of kinetic friction = 0.367

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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

Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then

Therefore, the coefficient of kinetic friction is approximately 0.0367

it is a negative acceleration