**AP Physics 1 Question 167: Answer and Explanation**

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**Question: 167**

**7.**

When a 0.20 kg block hangs at rest vertically from a spring of force constant 4 N/m, the spring stretches 0.50 m from its unstretched position, as shown in the figure. Subsequently, the block is stretched an additional 0.10 m and released such that it undergoes simple harmonic motion. What is the maximum kinetic energy of the block in its harmonic motion?

- A. 0.50 J
- B. 0.02 J
- C. 0.72 J
- D. 0.20 J

**Correct Answer:** B

**Explanation:**

**B**-You can look at this two ways. The hard way is to consider the spring energy gained and the gravitational energy lost in stretching the spring the additional 0.10 m separately. The block-earth system loses *mgh* = (0.20 kg)(10 N/kg)(0.10 m) = 0.20 J of gravitational energy; but the block-spring system gains ½*kx*_{2}^{2} – ½*kx*_{1}^{2} = {½(4 N/m)(0.60 m)^{2} – ½(4 N/m)(0.50 m)^{2}} = 0.22 J of spring energy. Thus, the net work done on the block in pulling it the additional 0.10 m is 0.02 J. That's what is converted into the block's maximum kinetic energy.

You can also look at it the easy way. With a vertical spring, consider the block-earth-spring system as a whole. Define the hanging equilibrium as the zero of the whole system's potential energy; then the potential energy of the whole system can be written as ½*kx*^{2}, where *x* is the distance from this hanging equilibrium position. That's ½(4 N/m)(0.10 m)^{2} = 0.02 J.