A marble moves along the x-axis. The potential-energy function is shown in the f

Question

A marble moves along the x-axis. The potential-energy function is shown in the figure (Figure 1) .

Part A

At which of the labeled x-coordinates is the force on the marble zero?

At which of the labeled -coordinates is the force on the marble zero?

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Part B

Which of the labeled x-coordinates is a position of stable equilibrium?

Which of the labeled -coordinates is a position of stable equilibrium?

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Part C

Which of the labeled x-coordinates is a position of unstable equilibrium?

Which of the labeled -coordinates is a position of unstable equilibrium?

Explanation / Answer

We know that the force F=-dU/dX.

PART A) ans b and d. When F=0, this implies dU/dx is minimum. at point b and d at point b and d F= -dU/dX. hence these point are equillibrium position.

PART B) ans b. because the stable equillibrium position at which potential energy is minimum. in the graph at point b potential energy is minimum.

PART C): ans d. from the graph at point d there is an increase in kinetic energy. adn hence marble easily runs away. hence it is unstable equillibrium.

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