A 5.00 kg package slides 1.50 m down a ramp that is inclined at 12 degrees below horizontal. The coefficient of kinetic friction is 0.310.
a- Calculate the work done by friction.
b- Calculate the work done by gravity.
c- Determine the net work done on package.
d- If thepackage started sliding with a speed of 2.2 m/s, find the final speed.
Fk = n * coefficient
W = Fd cos angle
K= i/2 mv^2
The Attempt at a Solution
n (Normal) = wcos 12= 47.9 N
Fk (Friction) = normal * coefficient = 14.8N
Work Fk = 1.5m * 14.8 N * cos 180 = -22.2 J
wsin12 * 1.5m * cos 0 = 15.3 J
W total = 15.3J - 22.2 J = -6.9 J
1/2 m v init ^2 - 6.9 J = 1/2 m vfinal ^2
12.1 J - 6.9 J = 2.5 kg v final ^2
v final = 1.44 m/s
I was feeling pretty good about this, but I showed it to someone in my school's math lab and he told me he thought it was all wrong. He said I need to break every force in x and y components. I don't see why I would. I would appreciate it if someone could tell me if I am doing this correctly.