Answer:
Value of coefficient of kinetic friction is 0.26 .
Explanation:
Given:
Mass of wooden crate, m = 12.5 kg.
Horizontal force to keep the block moving with constant velocity, F = 32.0 N.
Since, the block is moving with constant velocity.
So, net force experience by it is zero.
Therefore, fore of friction is equal to applied force.
Now, force of friction , [tex]F=\mu_kN[/tex] ( here [tex]\mu_k[/tex] is coefficient of kinetic friction and N is normal force)
Therefore, [tex]\mu_kN=\mu_k\times mg=\mu_k\times 12.5 \times 9.8=122.5\times \mu_k[/tex]
Now, both forces are equal.
[tex]122.5\times \mu_k=32\\\mu_k=\dfrac{32}{122.5}=0.26[/tex]
The value of coefficient of kinetic friction is 0.26 .
Hence, it is the required solution.
