0

Guest

Courses New

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

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

Question Image
Grade:11

2 Answers

Arun
25750 Points
6 years ago
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 
 
 
Regards
Arun (askIITians forum expert)
ABHINAV Shrivastav
13 Points
4 years ago
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

Think You Can Provide A Better Answer ?

Other Related Questions on General Physics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More