Question
A 2 kg block is pushed 2 m along a fixed horizontal surface by a horizontal force of 20 N. The coefficient of kinetic friction of the surface is µk= 0.40.
(a) How much work is done by the 20 N force?
(b) what is the work done by the friction force?
(c) What is the total work done by the net force on the object?
(a) How much work is done by the 20 N force?
(b) what is the work done by the friction force?
(c) What is the total work done by the net force on the object?
Answers
Wb = mg = 2kg * 9.8N/kg = 19.6N. = Weight of block.
Fb = 19.6N. @ 0deg. = Force of block.
Fp = 19.6sin(0) = 0 = Force parallel to
surface.
Fv = 19.6cos(0) = 19.6N. = Force perpendicular o surface.
Ff = u*Fv = 0.4 * 19.6N. = 7.84N. = Force of friction.
a. W = Fap * d = 20 * 2 = 40J.
b. W = Ff * d = 7.84 * 2 = 15.68J.
c. Fn = Fap - Fp - Ff,
Fn = 20 - 0 - 7.84 = 12.16N.
W = Fn * d =12.16 * 2 = 24.32J.
Fb = 19.6N. @ 0deg. = Force of block.
Fp = 19.6sin(0) = 0 = Force parallel to
surface.
Fv = 19.6cos(0) = 19.6N. = Force perpendicular o surface.
Ff = u*Fv = 0.4 * 19.6N. = 7.84N. = Force of friction.
a. W = Fap * d = 20 * 2 = 40J.
b. W = Ff * d = 7.84 * 2 = 15.68J.
c. Fn = Fap - Fp - Ff,
Fn = 20 - 0 - 7.84 = 12.16N.
W = Fn * d =12.16 * 2 = 24.32J.
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