Question

A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 39.0 cm/scm/s . What are:
a. The amplitude of the subsequent oscillations?
b. The block's speed at the point where x= 0.750 A?

Answer 1

Answer:

a

 $A = 0.081 \ m$

b

The value is  $u = 0.2569 \ m/s$

Explanation:

From the question we are told that

   The mass is  $m = 0.750 \ kg$

   The spring constant is  $k = 17.5 \ N/m$

    The instantaneous speed is  $v = 39.0 \ cm/s= 0.39 \ m/s$

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          $\frac{1}{2} * m * v^2 = \frac{1}{2} * k * A^2$

Here a is the amplitude of the subsequent oscillations

=>      $A = \sqrt{\frac{m * v^ 2 }{ k} }$

=>      $A = \sqrt{\frac{0.750 * 0.39 ^ 2 }{17.5} }$

=>       $A = 0.081 \ m$

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             $\frac{1}{2} * m * v^2 = \frac{1}{2} * k x^2 + \frac{1}{2} * m * u^2$

=>          $\frac{1}{2} * 0.750 * 0.39^2 = \frac{1}{2} * 17.5* 0.750(0.081 )^2 + \frac{1}{2} * 0.750 * u^2$

=>          $u = 0.2569 \ m/s$