Final velocity of a mass moving through friction

AI Thread Summary
The discussion revolves around calculating the work done by friction and the final velocity of a block being pulled at an angle. The work of friction was calculated to be -57.72 J, indicating that it exceeds the initial kinetic energy of the block, suggesting it would stop before reaching point b. The contributor then correctly accounted for the energy contributed by the pulling force, leading to a final velocity of 8.14 m/s at point b. There was also a note on the importance of including units consistently throughout the calculations. Overall, the calculations and reasoning provided a clear understanding of the dynamics involved in the scenario.
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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.
 
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That is the work of friction as it is
(-u)(normal force)(d)
u=0.3
d=4.0m
Normal force= (m*g)-(FsinΘ)
with
m=10kg
g=9.81m/s/s
Magnitude of force= 100N
Angle of force above horizontal= 30 degrees
 
Um, where are positions a and b? Without a diagram or clear description it's hard to proceed.

Presumably the 100 N force makes some contribution to the horizontal motion, not just alleviating some of the friction force. So how much energy does it contribute to the scenario?
 
Sorry point a is x=o, point b x=4m
gneill said:
Presumably the 100 N force makes some contribution to the horizontal motion, not just alleviating some of the friction force. So how much energy does it contribute to the scenario?
So:
FcosΘ*d+(0.5)(m)(v^2)+Work of friction=(0.5)(m)(vf^2)
100cos30*4+(0.5)(10kg)(3^2)-60=(0.5)(m)(v^2)
346.41+45-60=5v^2
v=8.14 m/s
 
Okay, the result looks good. Mind you, the lonely "kg" unit on the second line looks out of place without units being given for all the variables. So you may want to clean up the work before handing it in. Either include units for everything all the way though, use symbols for all values, or state that you're working with magnitudes and that units are ignored through the working (but definitely include the units on any final values!).
 
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