Question

A conservative force F(x) acts on a 2.0 kg particle that moves along an x axis....

A conservative force F(x) acts on a 2.0 kg particle that moves along an x axis. The potential energy U(x) associated with F(x) is graphed in the figure. When the particle is at x = 2.0 m, its velocity is -1.265 m/s.

What is F(2.)? (sign gives direction)

Between what positions on the left and right does the particle move? Left side?

Right side?

What is the particle's speed at x = 7.0 m

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Homework Answers

Answer #1


m = 2 kg, x = 2 m , v = - 1.265 m/s

(a) F = - dU/dx

slope of U and x graph gives F

F = U4 - U1/(4-1)

F = -( (-17.4) -(-2.7)/3)

F = 4.9 N , in +x direction

(b) At x = 2 m

U = - 8 J

K = (1/2)mv^2 = 0.5*2*1.265^2 = 1.6 J

E = U +K = - 8 +1.6 = - 6.4 J

This energy is strteched on graph.The range of possible values are between the intersection of E level with U grapph.

range is 1.5 m< x < 13.7 m

(c) at x = 0 m , U = -17 J

K = E -U = - 6.4 - (-17) = 10.6 J

(1/2)mv^2 = 10.6 J

0.5*2*v^2 =10.6

v = 3.26 m/s

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