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?

Expert's answer

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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