Asked by Adam
3813.75 Joules of work is being done on a 150 lb object while it is moved 75 ft across a horizontal plane, what is the final velocity if the coefficient of kinetic friction is 0.25 and the initial velocity is 1 ft / s?
Answers
Answered by
Henry
Mass = 150Lbs * 0.454kg/Lb = 68.1 kg.
d = 75Ft * 1m/3.3Ft = 22.73 m.
Work = F*d = F * 22.73 = 3813.75
Fn = 3813.75/22.73 = 167.8 N = Net force
a = Fn/mass = 167.8/68.1 = 2.46 m/s^2
Vo = 1Ft/s * 1m/3.3Ft = 0.303 m/s =
Initial velocity.
V^2=Vo^2 + 2a*d=0.303^2 + 2*2.46*22.73 =
111.9
V = 10.58 m/s = Final velocity.
d = 75Ft * 1m/3.3Ft = 22.73 m.
Work = F*d = F * 22.73 = 3813.75
Fn = 3813.75/22.73 = 167.8 N = Net force
a = Fn/mass = 167.8/68.1 = 2.46 m/s^2
Vo = 1Ft/s * 1m/3.3Ft = 0.303 m/s =
Initial velocity.
V^2=Vo^2 + 2a*d=0.303^2 + 2*2.46*22.73 =
111.9
V = 10.58 m/s = Final velocity.
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