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Old Oct 11th 2008, 03:11 PM   #1
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Work on an inclined Plane- Check if right please?

A 5.00 kg package slides 1.50 m down a long ramp that is inclined at 12 degrees below the horizontal . The coefficient of kinetic friction between ramp and package is 0.310.*

c) What is the net work done on the package?

Will it just be -22.3 J since the normal force and the weight is perpendicular to the x direction.*

d) If the package started sliding with a speed of 2.20 m/s, find it's final speed.

net work = change in KE
net work = 1/2 m (vf^2-vi^2)
= 1/2(5 kg) (vf^2 - 2.20^2)
Vf = 1.39 m/s (Right?)
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Old Oct 11th 2008, 04:27 PM   #2
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If I've understood your question correctly, the force of gravity isn't perpendicular to direction the package is moving. Therefore it's also doing work.

$\displaystyle W_{net}=W_R+W_G=- \mu mgcos(12) \cdot s+mgsin(12) \cdot s$

The normalforce is perpendicular to the direction it's moving.

Last edited by Aslak; Oct 11th 2008 at 05:05 PM.
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Old Oct 12th 2008, 10:35 AM   #3
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Exclamation

Originally Posted by Aslak View Post
If I've understood your question correctly, the force of gravity isn't perpendicular to direction the package is moving. Therefore it's also doing work.

$\displaystyle W_{net}=W_R+W_G=- \mu mgcos(12) \cdot s+mgsin(12) \cdot s$

The normalforce is perpendicular to the direction it's moving.
Could you say work done by weight:

Ww = W x X cos theta
Ww = mg x X cos theta
Ww = (5 kg) (9.80 m/s^2) (1.50 m) cos 12
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Old Oct 13th 2008, 12:16 PM   #4
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Wcos12 is the component of the weight perpendicular to displacement. Wsin12 is the component of the weight in the direction it's moving. Making a sketch often helps.

$\displaystyle W_w=mgsin(12) \cdot s=5,0 \cdot 9,8 \cdot sin(12) \cdot 1,5 \approx 15,3 \ J$
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