<p>in the lab you connect one end of a 15N/m spring to a ringstand and you hang

Question

<p>in the lab you connect one end of a 15N/m spring to a ringstand and you hang a 0.25 kg block from the other end of the spring. it hangs vertically in equilibrium<br />how far is the spring stretched from its relaxed length?<br /><br />suppose that you put your hand beneath the block and lift up on the block in order to partially support it .&#160;<br />it is still in equilibrium with the spring stretched 5 cm from its relaxed length?<br /><br />what is the force of your hand on the block?</p>
<p>&#160;</p>
<p>i tried to solve it&#160;<br /><br />|/f spring | = ks |s|&#160;<br />mg = 15 * s&#160;<br />9.8 * 0.25 = 15 s&#160;<br />s = 0.16 m&#160;<br /><br />is that right ?? or not ??<br />bcz he wants the ( far that the spring stretched from its relaxed length ) and he did not give me any length ?????</p>
<p>help me !!!!!!!!!!!!!</p>

Explanation / Answer

a)kx=mg x= mg/k = 0.25*9.8/15 = 0.163 m=16.3 cm b) F+kx=mg F = mg-kx = 0.25*9.8 - 15*0.05 =1.7N

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