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Grade: 12th pass
        
Please kindly see the attachment and solve question no 8 asap.....
Thanks for the help 😄😄😄😄😄😃😃
3 months ago

Answers : (1)

Vikas TU
9225 Points
							
Given data: Two block of masses 7 kg and 35 kg respectively. Block with mass of 35 kg is resting on the friction less ground and block of mass 7 kg is resting upon it as shown in the figure in the question.
The coefficient of friction, µs = 0.5
The coefficient of kinetic friction, µk = 0.4
Force, F = 100 N is applied on block of mass 7 kg.
Let us say block of mass 7 kg as “Block B” and block of mass 35 kg as “Block A”.
 
The limiting static frictional force on Block B, = µs * mass of B * g
           = 0.5 * 7 * 9.8 ………..[ taking g = 9.8 m/s² ]
           = 34.3 N
∴ 34.3 N
Hence, (limiting static frictional force on B)
∴ Since the static frictional force on B is less than the applied force of 100 N on it, so the block B will have some motion which will be then opposed by the kinetic friction acting upon it.
∴ force due to kinetic friction on B = µk * mass of B * g
                              = 0.4 * 7 * 9.8
                              = 27.44 N
Now, due to the opposed kinetic friction on B, block A will experience some motion. Since the ground on which A is resting has no friction therefore, force acting on A will equal to the force due to kinetic friction on B I.e., equal to 27.44 N.
We know,
  F = m * a
Therefore,  
Acceleration of block A = (force due to kinetic friction on B) / (mass of A)
                   = (µk * mass of B * g) / 35
                   = 27.44 / 35
                   = 0.784 m/s²
                   ≈ 0.8 m/s² 
3 months ago
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