Potential Energy Graph A conservative force F(x) acts on a 3.0 kg particle that

Question

Potential Energy Graph A conservative force F(x) acts on a 3.0 kg particle that moves along the x axis. The potential energy U(x) associated with F(x) is graphed in Fig. 10-46. When the particle is at x = 3.0 m, its velocity is -1.0 m/s. The "kinks" in the graph occur at (1, -2.8), (4, -17.2), and (8.5, -17.2); and the endpoint is at (15, -2).

Figure 10-46
(a) What are the magnitude and direction of F(x) at this position?
NDirection
negative x
positive x


(b) Between what limits of x does the particle move?
m (lower limit)
m (upper limit)
(c) What is its speed at x = 7.0 m?
m/s



DIAGRAM

http://img684.imageshack.us/i/0849.gif/

Explanation / Answer

a .
at point ( 15 , -2 ) ;

fx = - dU / dx ;

f x = - [ ( -2 - - 17.2 ) / ( 15 - 8.5 ) ]   = - 2.338 ;

so force mGNITUDE = 2.338 N ;

dirtection is negative x direction .

b .

(1, -2.8), (4, -17.2), ;

U - ( -2.8 ) = [ ( -17.2 - -2.8 ) / ( 4 -1 ) ] ( x - 1 ) ;

putting x = 3 ,

U ( 3 ) = -12.4 ;

Total energy = U ( 3 ) + 1 /2 * 3 * 1^2 = -10.9 ;

so its end point is where its U becomes -10.9 ;

U - ( -2.8 ) = [ ( -17.2 - -2.8 ) / ( 4 -1 ) ] ( x - 1 ) ;

-10.9 - ( -2.8 ) = [ ( -17.2 - -2.8 ) / ( 4 -1 ) ] ( x - 1 ) ;

or x = 2.5 ;

and

(8.5, -17.2); and the endpoint is at (15, -2).

U - ( -17.2 ) = [ ( -2 + 17.2 ) / ( 15 - 8.5 ) ] ( x - 8 .5 )     putting U = -10.9 ;

-10.9 - ( -17.2 ) = [ ( -2 + 17.2 ) / ( 15 - 8.5 ) ] ( x - 8 .5 )   

x = 11.194 ;

c .

totla energy = -10.9 ;

-10.9 = U(7) +k ;

-10.9 = -17.2 + K ;

K = 6.3 ;

0.5 * 3 v^2 = 6.3 ;

or v = 2.049 m /s <--ans

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