Question

A block is attached to one end of a spring with the other end of the spring fixed to a wall. The block is vibrating horizontally on a frictionless surface. If the mass of the block is 4.0 kg, the spring constant is k

Answer 1

Complete Question

A block is attached to one end of a spring with the other end of the spring fixed to a wall. The block is vibrating horizontally on a frictionless surface. If the mass of the block is 4.0 kg, the spring constant is k = 100 N/m, and the maximum distance of the block from the equilibrium position is 20 cm, what is the speed of the block at an instant when it is a distance of 16 cm from the equilibrium position?

Answer:

The velocity is $v = 0.6 \ m/s$

Explanation:

From the question we are told that

  The mass of the block is  m =  4.0 kg

  The spring constant is k =  100 N/m

  The maximum distance of the block from equilibrium position is  d = 20 cm =0.20 m

   The distance considered is  $d_k = 16 \ cm = 0.16 \ m$

Generally the maximum energy stored in the spring is mathematically represented as

      $E = \frac{1}{2} * k * d^2$

=>  $E = \frac{1}{2} *100 * 0.2^2$

=>  $E = 2.0 \ J$

Gnerally according to the law of energy conservation

   The energy maximum energy of  the spring = energy of  the spring at $d_k$ + energy of the block at $d_k$

Here energy of the block at $d_k$ is mathematically represented as

        $K_1 = \frac{1}{2} mv^2$

=>    $K_1 = \frac{1}{2} * 4* v^2$

=>    $K_1 = 2v^2$

Generally the energy of the spring at $d_k$ is mathematically represented as

      $E_2 =\frac{1}{2} * k * d_k^2$

=>    $E_2 =\frac{1}{2} * 100 * (0.16)^2$

=>    $E_2 =1.28 \ J$

So

       $2.0 = 1.28 + 2v^2$

=>    $v = 0.6 \ m/s$