What is the resulting acceleration of object B placed on object A?

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The discussion revolves around calculating the acceleration of object B, which is placed on object A, under the influence of a 100N force. The drag coefficient between the two objects is 0.3, with object A weighing 20kg and object B weighing 10kg. The resulting acceleration for object B is stated to be 7.06 m/s², but the method of calculation is questioned by participants. Key formulas mentioned include Newton's second law (ΣF = ma) and the kinematic equation for linear motion (s = 1/2 * a * t²). The thread seeks clarification on how the acceleration was derived and which block the calculations pertain to.
sedaw
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Object A placed on a floor frictionless and object B placed on object A.
The drag coefficient between A & B is 0.3
The length of A is 1m and the length of B is 30 cm .
a force of 100N workin on object B.
What is the time take to the edge of object B placed on the edge of object A ?


all i received its that acceleration of B a = 7.06



TNX
 
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ohh i forgat :
mA=20kg
mB=10kg
 
somebody ?

please
 
How did you calculated the acceleration?

Remember that in case of a linear motion: s =1/2*a*t^2.
 
\Sigma_F_=ma
f=ma
mBg*0.3=ma
 
sedaw said:
all i received its that acceleration of B a = 7.06
Did you calculate this or was it given to you? This is the acceleration of which block?
\Sigma_F_=ma
f=ma
mBg*0.3=ma
Which block is this? Please show your work.
 
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