Figure 8-48 shows a plot of potential energy U versus position x of a 0.88 kg pa

Question

Figure 8-48 shows a plot of potential energy U versus position x of a 0.88 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are U_1 = 25 J, U_2 = 40 J, and U_3 = 50 J. The particle is released at x = 4.5 m with an initial speed of 7.0 m/s, headed in the negative x direction. If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? What are the magnitude and direction of the force on the particle as it begins to move to the left of x = 4.0 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 7.0 m/s. If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? What are the magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m.

Explanation / Answer

mass of the partcle m=0.88 Kg

U1=25 J

U2=40 J

U3=50 J

a)

initially position of particel is,

x1=4.5 m

initila speed v1=7 m/sec

total energy T=U1+K1

T=25+1/2*m*v1^2

25+1/2*0.88*7^2

T=46.56

and

final position is x2=1m,

final speed is v2

by using engry realtion,

U1+K1=U2+K2

25+1/2*m*v1^2=40+1/2*m*v2^2

25+1/2*0.88*7^2=40+1/2*0.88*v2^2

==> v2=3.86 m/sec

b)

workdone to move from x=4 m is,

W=U2-U1

F*(x2-x1)=40-25

F*(4-2)=15

===> F=7.5 N (direction id positive)

c)

by using engry realtion,

Total energy is T=46.56 < U3=50 J

hence,

it cant reach point x=7m

and

particel reach if final K.E=0

hence particel will turn before reach the point x=7

let

particel will move from x=5 to x( here x is turning point)

then,

T=25+(U3-U2)*(x-5)

46.56=25+(50-25)*(x-5)

x=5.86 m

turning point x=5.86 m

d)

workdone to move from x=5 m is,

W=(U3-U2)

F*(x-x1)=50-40

F*(5-4)=10

===> F=10 N (direction is negative)

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