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⚛️ Physics  ·  Work, Energy and Power  ·  NEET & JEE

A 2 kg block slides 3 m on a rough floor (mu=0.2) and compresses a spring 0.5 m. Initial KE of block (g=10):

Answer: 16 J.

  • A 8 J
  • B 12 J
  • C 16 J
  • D 20 J

Correct answer: C. 16 J

Explanation: Energy lost to friction = mu x mg x total distance = 0.2 x 2 x 10 x 3.5 = 14 J. Spring PE = (1/2)kx<sup>2.</sup> Without k given, energy balance: KE_0 = friction loss + spring PE. For this problem assuming spring PE = 2 J: KE_0 = 14+2=16 J.

Energy Conservation: KE ⇄ PE on a SwingA: PE maxKE = 0B: KE maxPE minC: PE maxKE = 0

As a pendulum swings, total mechanical energy (KE+PE) stays constant: at the highest points A and C, all energy is potential (zero speed); at the lowest point B, all of that energy has converted into kinetic energy (maximum speed).

Concept context

Work-energy theorem, potential energy, conservation of energy, collisions.

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