MCQ
A body dropped from height $‘H’$ reaches the ground with a speed of $1.1 \sqrt {gH}$ . Calculate the work done by air friction? .............. $\mathrm{mgH}$
- A$0.395$
- ✓$-0.395$
- C$0.400$
- D$-0.400$
✓
Answer
Correct option: B.
$-0.395$
b
$\Delta \mathrm{K} . \mathrm{E} .$ of particle on reaching the ground
$\Delta \mathrm{K} . \mathrm{E} .$ of particle on reaching the ground
$=\frac{1}{2} \mathrm{m} \cdot(1.21) \mathrm{gh}=0.605 \mathrm{mgH}$
Now $\Delta \mathrm{KE}=\mathrm{mgH}-\mathrm{f}_{\mathrm{r}}^{\mathrm{air}}$
$0.605 \mathrm{mgH}=\mathrm{mgH}-\mathrm{f}_{\mathrm{r}}^{\mathrm{air}}$
$\Rightarrow \mathrm{f}_{\mathrm{x}}^{\mathrm{air}}=0.395 \mathrm{mgH}$
Now since body moves opposite to the direction of force of air friction.
$\Rightarrow \omega(\text { air friction })=-0.395 \mathrm{mgH}$
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