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A nonlinear spring is used to launch a toy car. The car is pushed against the spring, compressing the spring 2.5 cm. The force the spring exerts on the car is given by the equation F=−Kx2, where K=5000 Nm2. The potential energy stored in the spring when the car is pushed against it is most nearly:________________

Respuesta :

The potential energy stored in the spring when the car is pushed against it is most nearly is 0.026 J.

  • The calculation is as follows:

Non linear spring is

[tex]F = -kx^2 \rightarrow force[/tex]

The potential energy

[tex]\frac{dU}{dX} = -F\\\\U = \int\limits {-Fdx}\\\\U = + \int\limits^x_0 {kx^2dx}\\\\U = \frac{kx^3}{3} \\\\U = 5000 \div 3 (\frac{2.5}{100}^)2\\\\U = \frac{5000 \times 2.5\times 2.5\times 2.5}{3\times 100\times 100\times 100}[/tex]

= 0.026 J

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