Question

Spring lies on a horizontal table, and the left end of the spring is attached to a wall. the other end is connected to a box. the box is pulled to the right, stretching the spring. static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. the mass of the box is 0.80 kg, and the spring has a spring constant of 59 n/m. the coefficient of static friction between the box and the table on which it rests is s 0.74. how far can the spring be stretched from its unstrained position without the box moving when it is released?

Answer 1

The answer to your question is 20.18

Answer 2

The equilibrium condition allows finding the result for the maximum that the spring length is:

           x = 9.83 cm

Newton's second law establishes a relationship between the net force, the mass and the acceleration of the bodies, when the acceleration with zero is called the equilibrium condition.

               ∑ F = 0

A free-body diagram is a diagram of the forces without the details of the bodies, see attached.

Let's write the equilibrium condition for each axis.

x-axis

         $fr - f_e = 0 \\fr = f_e$

         

The friction and elastic forces have the form.

        $fr = \mu N \\f_e = - k x$

Let's substitute

        μ N = k x

y-axis

         N - W = 0

         N = W = m g

 

We look for the distance.

         μ mg = k x

         x = $\frac{\mu \ m \ g}{k}$  

Let's calculate.

         x =$\frac{0.74 \ 0.80 \ 9.8 }{59}$  

         x = 0.0983 m

         x - 9.83 cm

In conclusion using the equilibrium condition we can find the result for the maximum distance that the spring can be lengthened is:

           x = 9.83 cm

Learn more here:  brainly.com/question/20316059