The truck pulls a light weight of 1 * 10 ^ 3 kg with acceleration without an initial speed and covers

The truck pulls a light weight of 1 * 10 ^ 3 kg with acceleration without an initial speed and covers a distance of 0.4 km in 50 seconds. Determine the tension (mm) of the cable if its stiffness is 2 * 10 ^ 6 N / m. Young’s modulus is 200 GPa.

m = 1 * 10 ^ 3 kg.

V0 = 0 m / s.

t = 50 s.

S = 0.4 km = 400 m.

k = 2 * 10 ^ 6 N / m.

E = 200 GPa = 200 * 10 ^ 9 Pa.

Δl -?

Let us find the acceleration of a passenger car a from the formula for uniformly accelerated motion: S = V0 * t + a * t ^ 2/2. Since V0 = 0 m / s, then S = a * t ^ 2/2.

a = 2 * S / t ^ 2.

a = 2 * 400 m / (50 s) ^ 2 = 0.32 m / s2.

If we neglect the friction force, then the movement of the car occurs under the action of the cable tension force: m * a = Fcont.

Fcont / S = E * Δl / l0 – Hooke’s law.

Fcont = E * Δl * S / l0.

The value E * S / l0 is called the stiffness coefficient k: k = E * S / l0.

Fcont = k * Δl.

Δl = Fcont / k = m * a / k.

Δl = 1 * 10 ^ 3 kg * 0.32 m / s2 / 2 * 10 ^ 6 N / m = 0.00016 m = 0.16 mm.

Answer: the elongation of the transport cable is Δl = 0.16 mm.