Question

friction Q. A block of mass 5kg is on a rough horizontal surface and is at rest. Now a force of 24N is imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and g=9.8 m/s, then the acceleration of the block is 0.88m/s

Answer 1

As per the question, the mass of the block m = 5 kg.

The applied force F = 24 N.

It is given that acceleration due to gravity g = $9.8\ m/s^2$

The coefficient of kinetic friction $[\mu] =\ 0.4$

Hence, the frictional force acting on the block is calculated as -

              Frictional force f = $\mu mg$

                                          = $0.4\times 9.8\times 5\ N$

                                          = 19.6 N.

The net force acting on the particle $F_{net} = F-f$

                                                                  = 24 N - 19.6 N

                                                                  = 4.4 N.

We are asked to calculate the acceleration of the block.

From Newton's second law, we know that net force acting on particle is the product of mass with the acceleration .

      Mathematically  $F_{net} = ma$

                                  $a=\frac{F_{net}} {m}$

                                  $a=\frac{4.4}{5}\ m/s^2$

                                   $=0.88\ m/s^2$

Hence, the acceleration of the block is   $0.88\ m/s^2$