BPO 2023, Теория — Задача 1. Springs

• A. $\Delta U = mgR\sin\theta + \frac{1}{2}K(R\theta)^2$

  The official solution writes the work of F as the change in potential energy, with the weight written as mg and the spring constant K (not given in the statement).

• B1. $U(x) = \beta x + 0.5\alpha x^2$

• B2. 6 m/s

A. Solution:

In position A where the spring has its natural length we considered that the gravitational potential energy and the potential energy of the spring is zero.

The body moves very slowly. Consequently its kinetic energy does not change. (2 p)

If x is the strain of the spring at position Γ, then, for the spring-body system we have:

$\Delta U = U_2 - U_1 = U_g + U_s - 0 = mg\ R\sin\theta + 1/2\ K\ (R\ \theta)^2$ (4 p)

$U_g = mgh = mg\ (R\sin\theta)$ is the gravitational potential energy at position Γ (2 p)

$U_s = 1/2\ K\ x^2 = 1/2\ K\ (R\theta)^2$. the potential energy of the spring at position Γ (2 p)

B1. Solution:

(1 p)

By plotting the force F versus position x and calculating the area of the trapezoid formed from $x = 0$ to position x we will find

(2 p)

since at $x = 0$, $U_1 = 0$ (1 p)

then

(1 p)

B2. Solution:

We apply the principle of conservation of mechanical energy for the two positions $x_1 = 1.5$ m and $x_2 = 1.0$ m of the object and we have

(4 p)

(3 p)

For x=1.5 m

(0.5 p)

For x=1.0 m

(0.5 p)

Hence

(2 p)