Asked by Anonymous
A 295kg box is pulled at a constant speed by the little engine. The box moves a distance of 2.8 m across the horizontal surface. (μ=.20)
How much energy is transferred by the engine? (How much WORK is done?)
How much energy is transferred by the engine? (How much WORK is done?)
Answers
Answered by
Henry
m*g = 295 * 9.8 = 2891 N. = Wt. of box.
= Fn = Normal force.
Fs = u*Fn = 0.2 * 2891 = 578.2 N. =
Force of static friction.
Fe-Fs = m*a
Fe - 578.2 = m*0 = 0
Fe = 578.2 N. = Force of engine.
Energy=Work = Fe*d = 578.2 * 2.8=1619 J.
= Fn = Normal force.
Fs = u*Fn = 0.2 * 2891 = 578.2 N. =
Force of static friction.
Fe-Fs = m*a
Fe - 578.2 = m*0 = 0
Fe = 578.2 N. = Force of engine.
Energy=Work = Fe*d = 578.2 * 2.8=1619 J.
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