Chapter 08, Problem 077 A conservative force F(x) acts on a 1.9 kg particle that

Question

Chapter 08, Problem 077 A conservative force F(x) acts on a 1.9 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 -1.7 m/s. (a) What is F(x) at this position, including sign? Between what positions on the (b) eft and (c) right does the particle move? (d) What is its particle's speed at x7.0 m? x (m) 0 5 10 15 0 -5 -10 -15 -20 (a) Number (b) Number (c) Number (d) Number Units Units Units Units

Explanation / Answer

m= 1.9 kg

at x=2m

v= -1.7 m/s

a) we know the Force F = - dU/dx

dU = change in potential energy

dx= change in distance.

from x=1 m to x = 4m U is linear

hence f = - (-17.3 - (-2.7)) / (4-1) = + 14.4 / 3 = +4.8 N ( in x direction)

b) at x= 2 m

U= -8 J (from the graph)

kinetic energy = 1/2 * m * v2 = 2.89 J

so total energy = U + K = -5.11 J

so marking the value of E on graph, we get 2 points of intersection at x= 1.4 and 13.5 (approx)

so b) 1.4 m

c) 13.5 m

d) at x=7 m

U= -17 J

K = E-U = 11.89 J

so v = (2*K/m)1/2 = 4.3 m/s

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