Question

A 20 kg block rests on a rough horizontal table. A rope is attatched 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/s^2 . The coefficient is Kinetic force friction between the block and the table is ____( Round to the nearest hundredth )

Answer 1

Given :

Mass of block , M = 20 kg .

Force applied , F = 80 N .

Acceleration of block , $a=2.5\ m/s^2$ .

To Find :

The coefficient is Kinetic force friction between the block and the table .

Solution :

We know , Force equation on block is given by :

$F_{net }=F-\mu_k mg \\\\ma = F-\mu_k mg \\\\20\times 2.5 = 80 -\mu_k \times 20 \times 10\\\\\mu_k\times 200=30\\\\\mu_k=\dfrac{30}{200}\\\\\ mu_k=0.15$

Therefore , coefficient is Kinetic force friction between the block and the table is 0.15 .

Hence , this is the required solution .