
#1
Oct1807, 12:42 AM

P: 49

1. A conservative force F is directed along the horizontal x direction with F(x) = (2.0x
+4.0)N (where x is expressed in meters). It acts on a 5 kg mass that moves at constant velocity from x=1.0 m to x=5.0 m. The coefficient of kinetic friction of the mass with the surface is 0.1. a) Find the potential energy for the force. b) Calculate the work done by the conservative force and by the friction force. c) Calculate the change in potential energy. 2. (a)dU(r)/dr = F(r) (b) W=Fr (c) W= delta U 3. (a) U(x)=(x^2 +4x+C) where C is a constant is what I got after taking the integral of F(x) (b) I graphed the force versus the distance and took the area under the graph from x=1 to x=5. This gave me 40J for the work done by the conservative force. This is where i was confused. Wouldnt the Work done by friction have to be 40J since the object is in constant velocity. However, how would you calculate that? Wfriction=(static coeff)(mg)(r)=(0.1)(5 x 9.8)(4m)= about 20J is the way I initially thought of doing it but this leads to net work being done which isnt true for constant velocity problems right? (c)The change in potential energy should be 0 J because the object is in constant velocity and the net work done on the object is 0J. Or is the change in potential energy 40 J since if you plug in U(1) U(5) into the equation found in (a) you get 40 J Thanks in Advance!! 



#2
Oct1807, 01:00 AM

HW Helper
P: 4,125

I think the question is trying to "trick" you with the constant velocity part... the question doesn't explicitly state that there are no other forces acting on the object...
a) looks good. b) is just 20J I believe... just like you did (force of friction)*4m however c) I think is just 40J... using your potential energy formula... can you explain your reasoning for saying 0? 



#3
Oct1807, 01:04 AM

P: 49

oh okay so wait the object isnt in constant velocity?
(b) so if the W friction is 20 J, the work of the force can still be 40J? so the works dont have to be equal? (c) my reasoning for saying it was 0 was that i thought that net Work= change in potential energy. I thought that if the object was in constant velocity which means 0 net work and consequently the change in potential energy 0 J 



#4
Oct1807, 01:17 AM

HW Helper
P: 4,125

Work and Potential Energy! Please HelpThere's 2 ways to look at it... Net work done by all forces = change in kinetic energy net work done = 0. But net work done by nonconservative forces = change in kinetic energy + change in potential energy (for all the conservative forces involved). it could be that friction is the only nonconservative force... and everything else is conservative... in which case we can plug into this second equation: 20 = 0 + change in potential energy giving change in potential energy = 20J we could assume those other forces are nonconservative... which would mean that there's another nonconservative force doing 20J of work... 20 + (20) = 0 + change in potential energy. change in potential energy = 40J (here the only potential energy is due to the one given force). Maybe we're supposed to presume that anything other than friction is conservative... I'm not sure. 



#5
Oct1807, 01:24 AM

P: 49

ok that makes sense... this is a very poorly worded question since its not very clear. Thanks a lot for your help!




#6
Oct1807, 01:30 AM

P: 49

i was looking at (c) again and couldnt be just plug in 5 and 1 into the function U(x) calculatd in (a). I did U(5) U(1) and got 40J... How do we know that the Change in potential energy isnt +40J. Shouldnt it be positive since the work of the force is positive?




#7
Oct1807, 01:30 AM

HW Helper
P: 4,125





#8
Oct1807, 01:32 AM

HW Helper
P: 4,125

delta U = U (final)  U(initial) = U(5)  U(1) = 40J. However the work done by the conservative force involved is delta U = (40J) = 40J 



#9
Oct1807, 01:39 AM

P: 49

okay i get it now... thanks! i think i was just confused for a little bit since the work of gravity is Mg(H2H1) its equal to Wgrav= U(H1)U(H2). But this is because gravity points downward, opposite a raise in height.



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