Physics, asked by satyark5285, 1 year ago

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

Answers

Answered by SerenaBochenek
21

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

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