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clipperdude21
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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 isn't 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!
+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 isn't 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!
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