The trampoline surface is extended by 0.50 m when the child lands and its spring constant is 4600 N/m.The mass of the child is 32 kg.When the trampoline returns to its original length it causes the child to gain 450 J of kinetic energy.Calculate the energy transferred to the surroundings.Energy(surroundings) = _____ J
Question
The trampoline surface is extended by 0.50 m when the child lands and its spring constant is 4600 N/m.The mass of the child is 32 kg.When the trampoline returns to its original length it causes the child to gain 450 J of kinetic energy.Calculate the energy transferred to the surroundings.Energy(surroundings) = _____ J
Solution
To solve this problem, we first need to calculate the potential energy stored in the trampoline when it is extended by 0.50 m. The formula for potential energy stored in a spring is 0.5kx^2, where k is the spring constant and x is the extension of the spring.
- Calculate the potential energy stored in the trampoline: PE = 0.5 * k * x^2 PE = 0.5 * 4600 N/m * (0.50 m)^2 PE = 0.5 * 4600 * 0.25 PE = 575 J
Next, we know that when the trampoline returns to its original length, it causes the child to gain 450 J of kinetic energy. This energy must have come from the potential energy stored in the trampoline.
- Calculate the energy transferred to the child: Energy(child) = PE - KE Energy(child) = 575 J - 450 J Energy(child) = 125 J
Therefore, the energy transferred to the surroundings is 125 J.
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