A body of mass $1 ext{ kg}$ is thrown upwards with a velocity $20 ext{ ms}^{-1}$. It momentarily comes to re

Question

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

Accepted Answer

1. **Calculate Initial Mechanical Energy:** At the point of projection, the body has only kinetic energy ($K_i$) and zero potential energy ($U_i = 0$).
   $$ E_i = K_i = \frac{1}{2}mv^2 = \frac{1}{2} 	imes 1 	imes (20)^2 = 200	ext{ J} $$ [Class 11 Physics, Ch 5, Sec 5.4, Eq 5.5]
2. **Calculate Final Mechanical Energy:** At the maximum height ($h=18	ext{ m}$), the body is momentarily at rest ($v=0$), so kinetic energy is zero. It has gravitational potential energy ($U_f$).
   $$ E_f = U_f = mgh = 1 	imes 10 	imes 18 = 180	ext{ J} $$ [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.
   $$ 	ext{Energy Lost} = E_i - E_f = 200	ext{ J} - 180	ext{ J} = 20	ext{ J} $$ [Class 11 Physics, Ch 5, Sec 5.6, Eq 5.8b; Sec 5.8]

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