Question

Пружина с коэффициентом жесткости k=500H/m сжата силой F=100H. Определить работу А внешней силы , дополнительно сжимающей эту пружину ещё на L=2см. Трение в системе не учитывать .

Answer 1

Работа внешней силы пошла на увеличение потенциальной энергии пружины: A=E2-E1. E1 = k*x^2/2, где x=F/k. E2 = k(x+Δх)^2/2. Подставляем и получаем: A = k*(2x*Δх+Δх*Δх)/2 вот и все собственно.

Answer 2

Calculation of Work Done by External Force on the Spring

To determine the work done by the external force on the spring, we need to consider the formula for work:

Work (W) = Force (F) × Displacement (d) × cos(θ)

In this case, the force applied to the spring is 100 H and the additional compression of the spring is 2 cm (or 0.02 m). The angle between the force and the displacement is 0 degrees since the force is applied in the same direction as the displacement.

Using the given values, we can calculate the work done by the external force on the spring as follows:

Work (W) = Force (F) × Displacement (d) × cos(θ) W = 100 H × 0.02 m × cos(0°)

Since the cosine of 0 degrees is equal to 1, the equation simplifies to:

W = 100 H × 0.02 m × 1 W = 2 H

Therefore, the work done by the external force on the spring, when additionally compressed by 2 cm, is 2 H.

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