Question

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.

Answer 1

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

$u=\frac {v^{2}}{2gd}$

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

$u=\frac {1.2^{2}}{2*9.81*2}=0.0366972477064\approx 0.0367$

Therefore, the coefficient of kinetic friction is approximately 0.0367