<p><img class=\"mceBlock\" src=\"https://s3.amazonaws.com/answer-board-image/ff6

Question

<p><img class="mceBlock" src="https://s3.amazonaws.com/answer-board-image/ff6d192b-be94-4062-a1e1-d7ad5bf0c6fc.png" border="0" alt="uploaded image" />A spring-loaded toy gun is used to shoot a ball of mass m=1.5 kg straight up in the air, as shown in the figure. The spring has spring constant k=667 N/m. If the spring is compressed a distance of 25.0 centimeters from its equilibrium position y=0 and then released, the ball reaches a maximum height h_max (measured from the equilibrium position of the spring). There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y axis.</p>
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<td class="multipleChoiceOptions">a) Are each of these true or false?&#160;</td>
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<p>-Mechanical energy is conserved because no dissipative forces perform work on the ball.&#160;</p>
<p>-The forces of gravity and the spring have potential energies associated with them.&#160;</p>
<p>-No conservative forces act in this problem after the ball is released from the spring gun.</p>
<p>b)&#160;Find&#160;<img title="v_m" src="http://session.masteringphysics.com/render?var=v_m" alt="v_m" align="middle" />&#160;the muzzle velocity of the ball (i.e., the velocity of the ball at the spring's equilibrium position&#160;<img src="http://session.masteringphysics.com/render?tex=y%3D0" alt="y=0" align="middle" />).</p>
<p>c)&#160;Find the maximum height&#160;<img title="h_max" src="http://session.masteringphysics.com/render?var=h_max" alt="h_max" align="middle" />&#160;of the ball.</p>
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Explanation / Answer

A spring-loaded toy gun is used to shoot a ball of mass m =1.5 kg spring constant is k = 667 N /m If the spring is compressed a distance of  x = 25.0 cm a ) Applying conservation of energy 1/2 k x^2 + 1/2 mv^2 + mgx ( k / m ) x^2 = mv^2 + 2 mg x    v = ( (k/m)x^2 - 2gx )       = ( 667 N /m / 1.5 kg ) (25.0*10^-2 m )^2- 2 * 9.8 m/s^2 * 25.0*10^-2 m       = 4.78 m/s b ) Applying conservation of energy     1/2 m v^2 = mg h_max        h_max = v^2 / 2g                     = ( 4.78 m/s)^2 / 2 *9.8 m/s^2                      = 1.16m

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