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Asked Dec 12, 2019
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A block of mass 14.0 kg slides from rest down a frictionless 40.0° incline and is stopped by a strong spring with k = 2.30 ✕ 104 N/m.The block slides 3.00 m from the point of release to the point where it comes to rest against the spring. When the block comes to rest, how far has the spring been compressed?
 

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According to energy conservation law, potential energy of block at the top of inc...

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PE = PE, PEgire spring mgh=kx mg (Isine) = kx² (14.0kg)(9.8m/s³)((3.00m)(sin 40.0°)) = ÷(2.30 × 10ʻN/m)x² 264.6kg - m² /s° =(5.75 × 10° N/m)x² x- 264.6kg m²/s? 5.75 × 10°N/m = 0.214m

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