3. Block B is placed on a smooth surface Block A. is placed on a rough surface of Block B with coefficient of friction 0.60. The mass of A and B are 2kg and 4kg respectively. The frictional force between A and B (in N) is
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Answer:
8
Explanation:
Assume both blocks are accelerating towards left with acceleration a
a=f−22=20−f4 or f=8N
fmax=μmg=(0.6)(2)(10)=12N
So frictional force =8N
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Given: Mass of block A = 2kg
Mass of block B = 4kg
Coefficient of friction between block A and block B = 0.60
To find: Frictional force between A and B
Solution:
Since block A is kept on the surface of block B, therefore the mass of the system = 2+4 = 6 kg
The total force of the system = 18N
Therefore, acceleration will be 3 m/s².
Now we will check frictional force
Fr - 2 = mass×acceleration
Fr - 2 = 3×2
Fr = 8N
Since Fr< μmg
Thus, Fr = 8N
Therefore, the frictional force between A and B is 8N.
- The coefficient of friction is the measure of hindrance that the body experiences.
- The frictional force is the force that causes hindrance to the motion of the object, and it is always in the direction opposite to that of motion of the body.
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