How much energy is stored in a spring of constant 44.3 Nm-1 when it is compressed by 10.3 cm? Give your answer to one decimal place.

Question

BlackTom AI · Accepted Answer

To determine the energy stored in a compressed spring, we use the formula for elastic potential energy: U = (1/2) k x^2.

First, identify the given values: spring constant k = 44.3 N/m and compression x = 10.3 cm. Convert the displacement to meters: x = 10.3 cm = 0.103 m.

Compute x^2: (0.103 m)^2 = 0.010609 m^2.

Multiply by the spring constant: k x^2 = 44.3 × 0.010609 ≈ 0.46996 J.

Apply the 1/2 factor: U = (1/2) × 0.46996 J ≈ 0.23498 J.

Round to one decimal place as requested: 0.23498 J ≈ 0.2 J.

Therefore, the energy stored in the spring is about 0.2 joules. If you were to check the intermediate figures, you’d see the critical steps are ensuring correct unit conversion (cm to m), squaring the displacement, multiplying by k, and then applying the one-half factor followed by proper rounding.

How much energy is stored in a spring of constant 44.3 Nm-1 when it is compressed by 10.3 cm? Give your answer to one decimal place.数值题

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