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NEET PHYSICSEasy

A body of mass 1 kg1\text{ kg} is thrown upwards with a velocity 20 ms−120\text{ ms}^{-1}. It momentarily comes to rest after attaining a height of 18 m18\text{ m}. How much energy is lost due to air friction? (g=10 ms−2g=10\text{ ms}^{-2})

A

20 J

B

30 J

C

40 J

D

10 J

Step-by-Step Solution

  1. Calculate Initial Mechanical Energy: At the point of projection, the body has only kinetic energy (KiK_i) and zero potential energy (Ui=0U_i = 0). Ei=Ki=12mv2=12×1×(20)2=200 JE_i = K_i = \frac{1}{2}mv^2 = \frac{1}{2} \times 1 \times (20)^2 = 200\text{ J} [Class 11 Physics, Ch 5, Sec 5.4, Eq 5.5]
  2. Calculate Final Mechanical Energy: At the maximum height (h=18 mh=18\text{ m}), the body is momentarily at rest (v=0v=0), so kinetic energy is zero. It has gravitational potential energy (UfU_f). Ef=Uf=mgh=1×10×18=180 JE_f = U_f = mgh = 1 \times 10 \times 18 = 180\text{ J} [Class 11 Physics, Ch 5, Sec 5.7, Eq 5.11a]
  3. Apply Work-Energy Theorem: The change in mechanical energy is equal to the work done by the non-conservative force (air friction). The energy lost is the difference between initial and final energy. Energy Lost=Ei−Ef=200 J−180 J=20 J\text{Energy Lost} = E_i - E_f = 200\text{ J} - 180\text{ J} = 20\text{ J} [Class 11 Physics, Ch 5, Sec 5.6, Eq 5.8b; Sec 5.8]
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