Answer:
So we are applying 80 N on the block, 20kg in mass and it moves with the acceleration of 2.5 m/s/s
so we can find the force being applied on the block
F = ma
F = 20 X 2.5
F = 50 N
So only 50 N of the 80 N is being applied on the block.
Hence, the remaining 80-50 = 30 N force is being used to overcome friction
Force of Kinetic Friction = 30 N
Now, in this case the Normal Force will be equal to the Force applied by gravity(Mg)
So, Normal Force = 200 N
Coefficient of kinetic friction = Force of Kinetic Friction / Normal Force
Coefficient of kinetic friction = 30 / 200
Coefficient of Kinetic Friction = 0.15
The coefficient of kinetic friction between the block and the table is 0.153
The given parameters;
mass of the block, m = 20 kg
force applied to the rope, F = 80 N
acceleration of the block, a = 2.5 m/s²
The coefficient of kinetic friction between the block and the table is calculated as;
[tex]\mu _k = \frac{F_k}{N}[/tex]
where;
Fk is the frictional force
N is the normal reaction
The frictional force is calculated as follows
[tex]F_k = 80 - (20 \times 2.5)\\\\F_k = 80 - 50= 30 \ N[/tex]
The normal reaction force is calculated as follows;
N = mg
N = 20 x 9.8
N = 196 N
The coefficient of kinetic friction between the block and the table is;
[tex]\mu _k = \frac{F_k}{N}\\\\\mu_k = \frac{30}{196} = 0.153[/tex]
Thus, the coefficient of kinetic friction between the block and the table is 0.153
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