Friction versus Gravity (Work on an Incline)

AI Thread Summary
The discussion centers on the calculations of work done by gravity and friction on a 5kg block sliding down a 12-degree incline. The user questions the use of sin and cos of 348 degrees, which corresponds to 360° - 12°, in their calculations. It is emphasized that work is a scalar quantity and must accurately reflect the forces acting on the block. The gravitational force component acting parallel to the incline and the normal force for friction must be correctly identified for accurate calculations. The user seeks verification of their math and understanding of the angle used in the calculations.
Ronnin
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Can someone verify my calculations here? I have a 5kg block on an incline plane 12 degrees down angle. Mu k is .31 and slides a distance of 1.5m down the ramp.

Work gravity=5k(9.8 m/s^2)sin(348)(1.5m)=-15.29J
Work friction=.31(5k(9.8 m/s^2)cos(348)(1.5m)=22.29J

This can't be right, can it?
 
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Why sin or cos of 348, which looks like 360°-12°?

Work, like energy, is a scalar.

Work is given by the force applied over a distance.

One must assure the force is correctly characterized. Find the component of the gravitational force (weight) acting parallel with the incline. The friction is proportional (via friction factor) to the gravitational force component acting normal to the incline.

http://hyperphysics.phy-astr.gsu.edu/hbase/N2st.html
 
Astronuc said:
Why sin or cos of 348, which looks like 360°-12°?

Work, like energy, is a scalar.

Work is given by the force applied over a distance.

One must assure the force is correctly characterized. Find the component of the gravitational force (weight) acting parallel with the incline. The friction is proportional (via friction factor) to the gravitational force component acting normal to the incline.

http://hyperphysics.phy-astr.gsu.edu/hbase/N2st.html

I used 348 because the slope is downward from 0. Is any of my math wrong?
 
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