A single conservative force F(x) acts on a particle of mass m only along the x-a

Question

A single conservative force F(x) acts on a particle of mass m only along the x-axis. Its associated potential energy function is U(x) = 12x2 - 60x + 25 in Joules and it is graphed below. The x-axis is in meters. Suppose the particle is placed at a point of equilibrium and is given a brief push in the positive x-direction. The push gives the particle 75 Joules of initial kinetic energy. What are the approximate x-coordinates of the particle’s turning points?A.1.0 m and 4.0 m B.-0.4 m and 5.4 m C.-0.7 m and 5.75 mD.0.0 m and 5.0 m E.none of the above

Explanation / Answer

F = - dU/dx = 24 x - 60 = 0

x = 2.5 m

U(2.5) = 12(2.5)^2 - 60(2.5) + 25 = - 50 J

at turning Point, KE = 0

Applying energy conservation,

PEi + KEI = PEf + KEf

- 50 + 75 = (12x^2 - 60x + 25) + 0

12 x^2 - 60x = 0

x = 0 and 5 m
  
Ans(D)

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