Question

g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction between the block and the wall is 0.36. What is the minimum compression in the spring to prevent the block from slipping down?

Answer 1

Answer:

The minimum compression is  $x= 0.046m$

Explanation:

From the question we are told that

              The mass of the block is $m_b = 1.45 kg$

               The spring constant is  $k = 860 N/m$

               The coefficient of static friction is  $\mu = 0.36$

For the the block not slip it mean the sum of forces acting on the  horizontal axis is equal to the forces acting on the vertical axis

     Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

                   $F_y = m_b*g$

And the force acting on the horizontal axis is  force due to the spring which is mathematically represented as

                   $F_x = k *x * \mu$

where x is the minimum compression to keep the block from slipping

        Now equating this two formulas and making x the subject

                      $x = \frac{m_b * g}{k * \mu}$

substituting values we have

                     $x = \frac{1.45 * 9.8}{860 *0.36}$

                        $x= 0.046m$