posted by pakilina .
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?
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
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.