- #1
Jrlinton
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Homework Statement
A block of 10 kg is pulled by a force of 100 N at and angle 30 degrees above the horizonal through a distance of 4 meters. Kinetic friction has a coefficient of 0.3.
How much work does friction do in that distance?
If its speed is 3 m/s at point a, what is its speed at point b?
Homework Equations
The Attempt at a Solution
For the first question I calculated the work of friction to be :
-u(Fn)(d)
(-0.3)((10kg*9.81m/s/s)-(100sin30))(4m)
-57.72 J
Now for the second part I took the kinetic energy of the mass moving at 3 m/s and added the work done by friction and set that equal to the final kinetic energy.
(0.5)(m)(v^2)+Wf=(0.5)(m)(v^2)
(0.5)(10kg)((3m/s)^2)-57.72J=(0.5)(10kg)(v^2)
-12.72 J= 5v^2
This is where i got stuck as the work done by friction was greater than the energy of the system to begin with. So according to this the object stopped before it even reached point b.