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Grade: 11
A force of magnitude 2mg(1-ay) starts acting in the vertically upward direction on a body of mass m placed on earth’s surface, where y is the height of the object above the ground during the ascent and ‘a’ is a positive constant. The total height through which the body ascends is 
a) 2/a
b) 1/2a
c) 2/3 a
d) 1/a
2 years ago

Answers : (2)

23345 Points
Dear Devansh
In this case at max height eqillibrium will be there and in that condition,, 
Total downward force = Total upward force 
mg = 2mg ( 1 - ay ) 
1 = 2 ( 1 - ay ) 
1 = 2 - 2ay 
2ay = 2 - 1 
ay = 1/2 
y = 1/2a 
Arun (askIITians forum expert)
2 years ago
ABHINAV Shrivastav
13 Points
Initial and final K.E of the body is zero 
Applying work energy principal
Integration of Fnet.dy from  0 to h must equals to zero
Where fnet =F-mg
By putting value of fnet
We get [y-ay^2]from 0 to h =0
Finally we get h=1/a.hence option D is correct
7 months ago
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