A particle of mass 2 kg has a total energy of 1 J, and is acted on by only a con

Question

A particle of mass 2 kg has a total energy of 1 J, and is acted on by only a conservative force with spring constant k= 10 N/m and the associated potential energy function given by: U(x)= -mgx+ 1/2kx^2 a) Find the force, F(x), that is acting on the particle. b) Where is the equilibrium point for the particle? c) What is the force acting on the particle at the equilibrium point? d) What is the kinetic energy of the particle at the equilibrium point? e) What is the velocity of the particle at the equilibrium point? f) What are the turning points for the particle? g) Sketch the potential energy function.

Explanation / Answer

a) F(x) = - dU/dx = -mg + kx

b) At equilibrium point, F(xo) = 0

=> -mg + kxo = 0

=> equilibrium point, xo = mg/k = 2 * 9.81 / 10 = 1.96 m

c) Zero

d) Kinetic energy is the difference between total energy and potential energy of the particle.

KE = TE - PE = 1 - (-mgx + kx2/2)

At equilibrium point,

KEo = 1 + mgxo - kxo2/2

=> KEo = 1 + (2 * 9.81 * 1.96) - (10 * 1.962 / 2) = 20.25 J

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