AP Physics 1 Question 167: Answer and Explanation

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Question: 167


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


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 ½kx22 – ½kx12 = {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 ½kx2, where x is the distance from this hanging equilibrium position. That's ½(4 N/m)(0.10 m)2 = 0.02 J.