Question

An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the spring is 8.0 cm and the length of the spring when the object is in equilibrium is 5.9 cm. When the object is resting at its equilibrium position, it is given a sharp downward blow with a hammer so that its initial speed is 0.30 m/s. (a) To what maximum height above the floor does the object eventually rise

Answer 1

Answer:

The value is $A = 0.014 \ m$

Explanation:

From the question we are told that

    The mass of the object is  $m = 2.0 \ kg$

    The unstressed length of the string is  $l = 0.08 \ m$

    The length of the spring when it is  at equilibrium is  $l_e = 5.9 \ cm = 0.059 \ m$

      The initial speed (maximum speed)of the spring when given a downward blow $v = 0.30 \ m/s$

Generally the maximum speed  of the spring  is mathematically represented as

           $u = A * w$

Here A is maximum height above the floor (i.e the maximum amplitude)

            and $w$ is the angular frequency which is mathematically represented as

       $w = \sqrt{\frac{k}{m} }$

So

        $u = A * \sqrt{\frac{k}{m} }$

=>      $A = u * \sqrt{\frac{m}{k} }$

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

           $b = l -l_e$

=>         $b = 0.08 - 0.05 9$

=>         $b = 0.021 \ m$

Generally at equilibrium position the net force acting on the spring is  

            $k * b - mg = 0$

=>         $k * 0.021 - 2 * 9.8 = 0$

=>        $k = 933 \ N/m$

So

            $A = 0.30 * \sqrt{\frac{2}{933} }$

=>          $A = 0.014 \ m$