Question

A vertical ideal spring is mounted on the floor and has a spring constant of 116 N/m. A 0.60-kg block is placed on the spring in two different ways. (a) In one case, the block is placed on the spring and not released until it rests stationary on the spring in its equilibrium position. Determine the amount (magnitude only) by which the spring is compressed. (b) In a second situation, the block is released from rest immediately after being placed on the spring and falls downward until it comes to a momentary halt. Determine the amount (magnitude only) by which the spring is now compressed.

Answer 1

Answer:

a

 $x = 0.0507 \ m$

b

 $d = 0.1014 \ m$

Explanation:

From the question we are told that

   The spring constant is  $k = 116 \ N/m$

    The mass of the block is  $m = 0.60 \ kg$

Considering question a

  Generally the weight of the block is mathematically represented as

     $W = m * g$

=>  $W = 0.60 * 9.8$  

=>  $W = 5.88\ N$

Generally the force exerted on the spring is mathematically represented as

            $F = k * x$

At equilibrium

        $W = F$

=>     $k * x = 5.88$

=>     $116 * x = 5.88$

=>     $x = 0.0507 \ m$

Considering question b

  Generally the energy stored in the spring is mathematically represented as

     $E = \frac{1}{2} * k * d^2$

Now the potential energy of the block before it  is drooped is mathematically represented as  

      $PE = m * g * d$

   Generally from the law of energy conservation we have that

    $E = PE$

=>  $\frac{1}{2} * k * d^2 = m * g * d$

=>  $\frac{1}{2} * 116 * d = 0.60 * 9.8$

=>  $d = 0.1014 \ m$