25 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.doubt kya hai
The question itself , how do I do it?
Block B ka contact loose krwana hai mtlb normal zero krna hai
Jab A wla mass niche aayega to spring me kuch extension aayegi jisse block B par kuch upward force lgega or ek aisa x aayega kisi moment pr jisse Block B ka sara mass Spring force balance out kr dega or Block B just contact loose krni ki condition mai aajega
Further Block A ke niche jaane se Spring me jo extension ara hai usse spring potential energy store ho rhi hai
2 eqn banegi in dono cheeze se solve krke aajega minimum mass
let mass of B be M
Mg=kx
so we have to find at what x this is true
if a moves down by y metres
mgy=1/2ka^2 where a is extension in spring
how do we connect these two independent facts
we cant say agar ball falls a metres then spring should be elongated by a metres as well cuz spring is stiff , right ?
These are directly connected
If ball A moves by x the spring will extend by x , look at the figure once there is nothing in between A and spring
huh? but the spring doesnt work like string
springs are stiff right. how is it possible that if ball moves x the spring will be elongated by x
on second thoughts , it may be possible my bad
Look at the point of contact na at the end point of spring it is connected to string
warna toh ball wouldnt have moved in the first place
It is constrained to move the same distance as string
hm got your point
Or you can look at this in terms of energy
Jitna ball a apna potential energy loose krega utna energy hi to spring store krega right?
Mg=kx
mgx=1/2kx^2
mg=1/2kx
M=2m?
Yeah
book says M=m/2
m should be greater than or equal to M/2
oh i got it the other way
you hve written the same lol
👍
thanx man
Wlc
can u pls look at wpe2 @Real potato
can u tag me there
done
+solved @Real potato
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