Final velocity of a mass moving through friction

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Homework Help Overview

The discussion revolves around a physics problem involving a block being pulled by a force while experiencing kinetic friction. The participants are analyzing the work done by friction and the final velocity of the block after moving a specified distance.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of work done by friction and its implications on the kinetic energy of the block. There are attempts to clarify the contributions of the applied force to the block's motion and questions about the positions of points a and b.

Discussion Status

Some participants have provided calculations and interpretations of the work done by friction and the energy contributions from the applied force. There is an acknowledgment of the need for clarity regarding the positions in the problem, and one participant has suggested a revised approach to calculate the final velocity.

Contextual Notes

Participants note the absence of a diagram or clear descriptions of the positions involved, which may affect the analysis. There is also a mention of the need for consistent unit usage in the calculations presented.

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.
 
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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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