Answer to Question #104977 in Mechanics | Relativity for Ceaser

Question #104977
A block of mass 12.0 kg slides from rest down a friction- less 35.0° incline and is stopped by a strong spring with k 5 3.00 3 104 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?
1
Expert's answer
2020-03-10T11:27:13-0400

According to the law of conservation of energy


kx22=mgLsin35°\frac{kx^2}{2}=mgL\sin35°


x=2mgLsin35°k=2129.83sin35°31040.12mx=\sqrt{\frac{2mgL\sin35°}{k}}=\sqrt{\frac{2\cdot 12\cdot 9.8\cdot 3\cdot \sin35°}{3\cdot 10^{4}}}\approx0.12m








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