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What is the magnitude of the average contact force exerted on the leg? Physics Help?

0 votes

A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 milliseconds from an initial speed of 4.25 m / s., assuming the total mass of the hand and the forearm to be 1.65 kg?

asked Nov 1, 2014 in PHYSICS by anonymous
reshown Nov 1, 2014 by bradely

1 Answer

0 votes

The total mass of the hand and forearm = 1.65 kg.

Initial Speed = 4.25 m/sec

The hand comes to rest then final Velocity = 0 m/sec.

Time Duration = 2.65 msec = 2.65 * 10^-3 sec.

Force = Mass * Accerlation

Accerlation = Change in velocity / change in time.

Accerlation = (Final Velocity - Initial Velocity) / Time interval

Accerlation = (0 - 4.25) / 2.65 * 10^-3

Accerlation = -1603.77 m/sec²

Therefore Force = Mass * Accerlation.

F = 1.65 kg * -1603 m/sec²

F = -2646 N.

Force = -2.646 kN.

Therefore the average contact force exerted on the leg is -2.646 kN.

answered Nov 1, 2014 by dozey Mentor
edited Nov 1, 2014 by dozey

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