helpbto solve this problem please, question no.14
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selena17:
please help
its answer should be 2/3 F
very urgent
mass of H = 2 x mass of G
if m(H) = 2m then m(G) = m
total force = 2m + m x a = 3ma
force on h = 2ma/3ma = 2/3
Answers
Answered by
1
Let acceleration be a m/s^2
Let mass of G be m kg
therefore, mass of H be 2m kg
Now,
Total force will be
F=(total mass) * acceleration
F=(m+2m) * a
F=(3m)*a
F=3ma ----------------(1)
Now, Let Force act on H By G be T
therefore, Here we will take net force act on H by G and that will be
F - T = (mass of G) * a
F - T = ma
F - ma = T
From equation 1
3ma - ma = T
T = 2ma -----------------(2)
Now,As we have to find it in terms of F
In equation 2, Divide and multiply 3.
T =(2ma *3)/3
By just arranging the terms
T = (3ma) * 2/3
By equation 1
T = (2F)/3
Answer:- option 1
Let mass of G be m kg
therefore, mass of H be 2m kg
Now,
Total force will be
F=(total mass) * acceleration
F=(m+2m) * a
F=(3m)*a
F=3ma ----------------(1)
Now, Let Force act on H By G be T
therefore, Here we will take net force act on H by G and that will be
F - T = (mass of G) * a
F - T = ma
F - ma = T
From equation 1
3ma - ma = T
T = 2ma -----------------(2)
Now,As we have to find it in terms of F
In equation 2, Divide and multiply 3.
T =(2ma *3)/3
By just arranging the terms
T = (3ma) * 2/3
By equation 1
T = (2F)/3
Answer:- option 1
thankyou soooo much
for answering
My pleasure
Answered by
2
If the boxes are in contact. Then they must have a common acceleration. Let it be equal to and. Masses of block b m & 2m then F=3ma fron here a=F/3m . force applied on larger block is only due to smaller block(leave normal rxⁿ by ground as body is moving horizontally.) Then Force on big block =mass×acclⁿ =2m×F/3m =2F/3
I always have the shortest and the best solⁿ. Isn't it akshara?
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