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

Question

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

Accepted Answer

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 ($KE_i$):** 
    $$KE_i = \frac{1}{2}mv^2 = \frac{1}{2} 	imes 1 	ext{ kg} 	imes (20 	ext{ ms}^{-1})^2 = 200 	ext{ J}$$
2.  **Final Potential Energy ($PE_f$) at 18 m:**
    $$PE_f = mgh = 1 	ext{ kg} 	imes 10 	ext{ ms}^{-2} 	imes 18 	ext{ m} = 180 	ext{ J}$$
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) . 
    $$	ext{Energy Lost} = KE_{initial} - PE_{final}$$
    $$	ext{Energy Lost} = 200 	ext{ J} - 180 	ext{ J} = 20 	ext{ J}$$

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

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