A bar weighing 490 g lies on a horizontal plane and is connected to the vertical wall by an undeformed spring

A bar weighing 490 g lies on a horizontal plane and is connected to the vertical wall by an undeformed spring. The axis of the spring is horizontal, the spring rate is 180 N / m. A bullet weighing 10 g gets stuck in the block, the velocity of which is parallel to the axis of the spring, as a result of which the spring is compressed by 10 cm. What is the initial velocity of the bullet? The coefficient of friction between the bar and the floor is 0.2. g = 10 m / s2.

mb = 490 g = 0.49 kg.

k = 180 N / m.

mp = 10 g = 0.01 kg.

x = 10 cm = 0.1 m.

μ = 0.2.

g = 10 m / s2.

V -?

To solve this problem, we will use the energy conservation law.

The kinetic energy of the bullet Ekp, when it hits the bar, is converted into the potential energy of the compressed spring En and overcoming the friction force between the bar and the surface.

Ekp = En + A, where A is the work of the friction force.

Ekp = mp * V ^ 2/2.

En = k * x ^ 2/2.

A = Ftr * x.

Ftr = μ * (mb + mp) * g.

A = μ * (mb + mp) * g * x.

mp * V ^ 2/2 = k * x ^ 2/2 + μ * (mb + mp) * g * x.

V2 = k * x ^ 2 / mp + 2 * μ * (mb + mp) * g * x / mp.

V = √ (k * x ^ 2 / mp + 2 * μ * (mb + mp) * g * x / mp).

V = √ (180 N / m * (0.1 m) ^ 2 / 0.01 kg + 2 * 0.2 * (0.49 kg + 0.01 kg) * 10 m / s2 * 0.1 m / 0.01 kg) = 14.1 m / s.

Answer: the bullet flew at a speed of V = 14.1 m / s.