A girl and a boy each exert a force to move a crate horizontally a distance of 13m at a constant speed. The girl's applied forced is 75N [22° below the horizontal]. The tension in the rope pulled by the boy is 75N [32° above the horizontal]. Determine the work done by the boy and the girl. Determine the force of friction and the work done by friction. If the coefficient of kinetic friction between the box and the floor is 0.5, what is the mass of the crate? Answer: W girl = Ft As = (75cos 22° N)(13m) - 9.04x10° J girl x W =(75cos 32°N)(13m) 8.27×10°J %3D boy=FhoyAs boy.x Constant speed means a=0, which means net work done is also zero. As a result, W, =-(9.04x10° +8.27×10)J =-1.73×10'J. W, = F,As=F, =133N. F, = µF Fy = 266N. Fy = mg -72sin 22° – 75 sin 32° →m=34kg %3D Sourse: modified from Nelson, Physics 12, 2002, p.183, Section 4.1, #5

A girl and a boy each exert a force to move a crate horizontally a distance of 13m at a constant speed. The girl's applied forced is 75N [22° below the horizontal]. The tension in the rope pulled by the boy is 75N [32° above the horizontal]. Determine the work done by the boy and the girl. Determine the force of friction and the work done by friction. If the coefficient of kinetic friction between the box and the floor is 0.5, what is the mass of the crate? Answer: W girl = Ft As = (75cos 22° N)(13m) - 9.04x10° J girl x W =(75cos 32°N)(13m) 8.27×10°J %3D boy=FhoyAs boy.x Constant speed means a=0, which means net work done is also zero. As a result, W, =-(9.04x10° +8.27×10)J =-1.73×10'J. W, = F,As=F, =133N. F, = µF Fy = 266N. Fy = mg -72sin 22° – 75 sin 32° →m=34kg %3D Sourse: modified from Nelson, Physics 12, 2002, p.183, Section 4.1, #5