Calculate the work done by friction of package

AI Thread Summary
The discussion revolves around calculating the work done by friction, gravity, and the net work on a 5.00 kg package sliding down a 12-degree ramp with a coefficient of kinetic friction of 0.310. The work done by friction is calculated as -22.2 J, while the work done by gravity is 15.3 J, resulting in a net work of -6.9 J. The final speed of the package, starting from an initial speed of 2.2 m/s, is determined to be 1.44 m/s after accounting for the net work. A participant expresses concern over the accuracy of their calculations, suggesting that breaking forces into components might be necessary, but they receive confirmation that their approach is correct. The calculations and results are validated, leading to a satisfactory conclusion.
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Homework Statement



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.

Homework Equations



Fk = n * coefficient
W = Fd cos angle
K= i/2 mv^2

The Attempt at a Solution



a-
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

b-
wsin12 * 1.5m * cos 0 = 15.3 J

c-
W total = 15.3J - 22.2 J = -6.9 J

d-
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.
 
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You are correct.
 
Alright! Thank you.
 
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