A) 2.8 mgH
B) \[-1.3\,mgH\]
C) 1.3 mgH
D) \[-0.28\,mgH\]
Correct Answer: D
Solution :
The forces acting on the body are (i) force of gravity, (ii) air-friction According to work-energy theorem, total work done on the body = Gain in KE \[W=\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}m{{(1.2\sqrt{gH})}^{2}}\] \[=0.72mgH\] As work done by gravity, \[{{W}_{1}}=mgH\] \[\therefore \]Work done by friction, \[{{W}_{2}}=W-{{W}_{1}}\] \[=0.22mgH-mgH\] \[=-0.28\text{ }mgH\]You need to login to perform this action.
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