They both attract things.
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The springs stored energy is transferred to the cube as kinetic energy and then by the slop the KE is converted to height energy.
<span>0.5 . k . x^2 = 0.5 . m . v^2 = m . g . ∆h </span>
<span>0.5 . 50 . (0.1^2) = 0.05 . 9.8 . ∆h </span>
<span>∆h = 0.51 m = 51 cm </span>
<span>This is the height gained </span>
<span>Distance along the slope = ∆h / sin 60 = 0.589 = 59 cm </span>
<span>In the second case, the stored spring energy is converted into height energy AND frictional heat energy. </span>
<span>The height energy is m . g . d sin 60 where d is the distance the cube moves along the slope. </span>
<span>The Frictional energy converted is F . d </span>
<span>F ( the frictional force ) = µ . N </span>
<span>N ( the reaction to the component of the gravity force perpendicular to the surface of the slope ) = m . g . cos60 </span>
<span>Total energy converted </span>
<span>0.5 . k . x^2 = (m . g . dsin60) + (µ . m . g . cos60 . d ) </span>
<span>Solve for d </span>
<span>d = 0.528 = 53 cm</span>
Kinetic energy than parked
It might be 4.0 or 2.22344 seconds as velocity speed
