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wel
2 days ago
5

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 suitca

se slides 2.0 m before stopping. Use work and energy to find the suitcase’s coefficient of kinetic friction on the floor.
Physics
1 answer:
ValentinkaMS [3K]2 days ago
7 0

Answer:

0.0367

Explanation:

The decrease in kinetic energy translates into the work carried out by friction.

Since the formula for kinetic energy is

KE=0.5mv^{2}

The work done by friction is characterized as

W= umgd

Where m represents the suitcase's mass, v signifies its velocity, g is the acceleration of gravity, d is the perpendicular distance where the force is applied, and u denotes the kinetic friction coefficient.

Rearranging for u allows us to deduce that

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

Substituting v with 1.2 m/s, d with 2m, and using g as 9.81 m/s² yields

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

Consequently, the coefficient of kinetic friction is roughly 0.0367

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