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

Question

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 -1.8 m/s. (a) What is F(x) at this position, including sign? Between what positions on the (b) left and (c) right does the particle move? (d) What is its particle's speed at x = 7.0 m? (a) Number Units (b) Number Units (c) Number Units (d) Number Units Click If you would like to Show Work for this question: Open Show Work

Explanation / Answer

From the graph,

The kink occur at (1, -2.8), (4, -17.2), and (8.5, -17.2) and the endpoint is at (15, -2)

(a)

We kknow that,

F(x) = -dU / dx

F = -17.2 - (2.8) / (4 - 1)

F = +4.8 N

(b)

The particle move b/w 1.5 to 13.5 m

x = 1.5 m (left)

(c)

x' = 13.5 m (right)

(d)

Total energy of particle when particle at x = 2 m

TE = (1/2)mv^2 + PE

TE = (1/2)*2*(-1.8)^2 - 7.7

TE = -4.376

From energy conservation,

KE at at x = 7 m,

KE = -4.376 - (-17.5)

KE = 13.124 J

13.124 = (1/2)*2*v^2

v = 3.62 m/s

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