A) \[{{t}_{1}}<{{t}_{2}};{{W}_{1}}>{{W}_{2}}\]
B) \[{{t}_{1}}>{{t}_{2}};{{W}_{1}}<{{W}_{2}}\]
C) \[{{t}_{1}}={{t}_{2}};{{W}_{1}}={{W}_{2}}\]
D) \[{{t}_{1}}>{{t}_{2}};{{W}_{1}}={{W}_{2}}\]
Correct Answer: D
Solution :
The average velocity in the first half of the distance \[=\frac{0+v}{2}=\frac{v}{2};\] while in the second half, the average velocity is v. therefore,\[{{t}_{1}}>{{t}_{2}}\]. The work done against gravity in both halves =\[mgh=mgL/2\]. Hence the correct choice is [D].You need to login to perform this action.
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