Question

A block of mass 27.00 kg sits on a horizontal surface with, coefficient of kinetic

Answer 1

Answer:

The force is  $F = 172 \ N$

Explanation:

From the question we are told that

    The  mass of the block is  $m_b = 27.0 \ kg$

     The  coefficient of  static friction is  $\mu_s = 0.65$

     The coefficient of kinetic friction is  $\mu_k = 0.50$

The  normal force acting on the block is  

      $N = m * g$

substituting values

     $N = 27 * 9.8$

     $N = 294.6 \ N$

Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as

        $F_f = \mu_s * N$

substituting values

        $F_f = 0.65 * 264.6$

        $F_f = 172 \ N$

Now for this  block to move the force require is  equal to $F_f$ i.e

       $F= F_f$

=>    $F = 172 \ N$