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Answer :
Final answer:
The question pertains to the conservation of energy concept in physics. The potential energy stored in the compressed spring is transformed into gravitational potential energy when the bait is launched up the incline. By equating these two energies, we can solve for the maximum height reached by the bait.
Explanation:
Your question is related to the physics concept of conservation of energy. The situation involves a spring with a constant k-125/'m which shoots a small 0.100kg object (in this case, a bait) up a frictionless incline after being compressed by 0.750m. Our task is to find out the maximum height (h) that the object reaches.
When a spring is compressed, it stores potential energy equal to 0.5*k*x^2, where 'k' is the spring constant and 'x' is the compressed distance. In this scenario, the spring catapults the bait up an incline, converting all the potential energy into gravitational potential energy at the maximum height. This can be represented as m*g*h, where 'm' is the mass of the bait, 'g' is the acceleration due to gravity (about 9.8 m/s^2 on Earth), and 'h' is the height.
By setting these two equalities, we get 0.5*k*x^2 = m*g*h. Plugging in the known values, we find h = (0.5*125*0.750^2) / (0.100*9.8). Calculate that, and you will get the maximum height reached by the bait.
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