A 17.8 kg block is dragged over a rough, horizontal surface by a constant force of 72.8 newtons acting at an angle of 31.3 degrees above the horizontal. the block is displaced 64.4 m and the coefficient of kinetic friction is 0.239. find the work done by the 72.8 n force. answer in units of J

Answer 1

Wb = M*g = 17.8 * 9.8 = 174.4 N.

Fn = Mg-Fap*sin31.3 = 174.4-72.8*sin31.3
= 136.6 N. = Normal force.

Fk = u*Fn = 0.239*136.6 = 32.7 N.

Work = (Fap*Cos31.3-Fk) * d =
(72.8-32.7) * 64.4 =