The figure below shows a plot of potential energy U versus position x of a 1.08

Question

The figure below shows a plot of potential energy U versus position x of a 1.08 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are U1 = 5 J, U2 = 30 J, and U3 = 50 J.

The particle is released atx=4.5m with an initial speed of6.0m/s, headed in the negativexdirection.

(a) If the particle can reachx= 1.0 m, what is its speed there, and if it cannot, what is its turning point?

(b) What are the magnitude and direction of the force on the particle as it begins to move to the left ofx= 4.0 m?

Suppose, instead, the particle is headed in the positivexdirection when it is released atx=4.5m at speed6.0m/s.

(c) If the particle can reachx= 7.0 m, what is its speed there, and if it cannot, what is its turning point?

(d) What are the magnitude and direction of the force on the particle as it begins to move to the right ofx= 5.0 m?

Explanation / Answer

total energy = 5+ 0.5*1.08*36 = 24.44 J but given u3 = 50 J not impossible

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