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NEET PHYSICSMedium

A body of mass 1 kg is thrown upwards with a velocity 20 ms1^{-1}. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (Take g = 10 ms2^{-2})

A

20 J

B

30 J

C

40 J

D

10 J

Step-by-Step Solution

To find the energy lost due to air friction, we apply the Law of Conservation of Energy, which states that the total initial energy of the body must equal the total final energy plus any energy dissipated by non-conservative forces like friction .

  1. Initial Kinetic Energy (KEiKE_i): KEi=12mv2=12×1 kg×(20 ms1)2=200 JKE_i = \frac{1}{2}mv^2 = \frac{1}{2} \times 1 \text{ kg} \times (20 \text{ ms}^{-1})^2 = 200 \text{ J}
  2. Final Potential Energy (PEfPE_f) at 18 m: PEf=mgh=1 kg×10 ms2×18 m=180 JPE_f = mgh = 1 \text{ kg} \times 10 \text{ ms}^{-2} \times 18 \text{ m} = 180 \text{ J}
  3. Energy Balance: Initially, the body has 200 J of energy. At its highest point, it only possesses 180 J of potential energy (since its velocity, and thus its kinetic energy, is zero) . Energy Lost=KEinitialPEfinal\text{Energy Lost} = KE_{initial} - PE_{final} Energy Lost=200 J180 J=20 J\text{Energy Lost} = 200 \text{ J} - 180 \text{ J} = 20 \text{ J}

Thus, 20 J of energy was dissipated as heat due to air friction .

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