A 20 kg block rests on a rough horizontal table. A rope is attached to the block and is pulled with a force of 80 N to the
left. As a result, the block accelerates at 2.5 m/s2. The
coefficient of kinetic friction uk between the block and the table
is

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;

where;

Fk is the frictional force

N is the normal reaction

The frictional force is calculated as follows

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;

Thus, the coefficient of kinetic friction between the block and the table is 0.153

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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

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

unbalanced force

explanation: