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Working with projectile motion.

2. A 3.0 kg object subject to a restoring force F is undergoing simple harmonic motion with small amplitude.

(See attachment)

The potential energy U of the object as a function of distance x from its equilibrium position is shown above.
This particular object has a total energy E of 0.4 J.

(a) What is the object's potential energy when its displacement is +4cm from its equilibrium position?

(b) What is the farthest the object moves along the x-axis in the positive direction? Explain your reasoning.

(c) Determine the object's kinetic energy when its displacement is -7 cm.

(d) What is the object's speed at x = 0?

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The potential energy U of the object as a function of distance x from its equilibrium position is shown above.
This particular object has a total energy E of 0.4 J.

The key point in solving this problem is that, the total energy of the object remains constant. That is, the sum of Kinetic Energy (due to the motion) and the Potential Energy (due to the displacement from the mean position) is equal to its total energy = 0.4 Joule. When the body moves, KE and PE changes. But the total remains the same.

If a body of mass m is moving with a velocity v, its Kinetic Energy (hereafter KE) is given by the familiar expression KE = (½) mv2

The body can be assumed to be executing simple harmonic motion. Let the restoring force (force pull it towards the point 0 in the figure) be given by F = -k x, where x is the displacement.

The potential energy is given by, PE = ...

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Complete answers as a Word attachment. Esay and step by step explanations are provided.

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