1. A box mass m=1 kg is pushed up an incline of an angle theta=30 degrees that has a coefficient of kinetic friction u_k=.5. Find the velocity of the object after it pushed for d=2m by a force of magnitude F=11N directed upward alone the incline. W_g=m*g*d*cos(90+theta) W_constant force=F*d*cos(theta) F_f=m*g*u_k F_n=m*g W(which is the sum of the work done on the box)=1/2*m*v^2 First off I solved for the W_g, which is just straight substitution W_g=1*9.8*2*cos(120)=-9.8. After that I solved for the constant force W_constant force= 11[N]*2*cos(30)=19.05. Now this is the part where I try to determine the work that friction affects the box which is F_f=1*9.8*.5=4.9 and I am not for sure if I need to solve for the work done by friction W_f=4.9*2*cos(30) or not and i know the normal force is 0 since F_n is perpendicular to the displacement .... I know the answer is 3.2 [M/s] but I am off by a bit every time. Please help!!!