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katovenus [111]
3 years ago
14

A spring is stretched to a displacement of 3.4 m from equilibrium. Then the spring is released and allowed to recoil to a displa

cement of 1.9 m from equilibrium. The spring constant is 11 N/m. What best describes the work involved as the spring recoils?
Physics
1 answer:
Stells [14]3 years ago
5 0
Answer to A spring<span> is </span>stretched<span> to a </span>displacement<span> of </span>3.4 m<span> from </span>equilibrium<span>. </span>Then<span> the </span>spring<span> is</span>released<span> and ... </span>Then<span> the </span>spring<span> is </span>released<span> and </span>allowed<span> to </span>recoil<span> to a </span>displacement<span> of </span>1.9 m<span> from</span>equilibrium<span>. The </span>spring constant<span> is </span>11 N/m<span>. What </span>best describes<span> the </span>work involved<span> as the </span>spring recoils<span>? A)87 J of </span>work<span> is performed ...</span>
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All of Dina's potential energy Ep is converted into kinetic energy Ek so Ep=Ek, where Ep=m*g*h and Ek=(1/2)*m*v². m is the mass of Dina, h is the height of ski slope, g=9.8 m/s² and v is the maximal velocity. 

So we solve for v:

m*g*h=(1/2)*m*v², masses cancel out,

g*h=(1/2)*v², we multiply by 2,

2*g*h=v² and take the square root to get v

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v=9.9 m/s. 

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