Find the elongation of the towing cable with a stiffness of 0.01 MN / m when towing a car with a mass of 2

Find the elongation of the towing cable with a stiffness of 0.01 MN / m when towing a car with a mass of 2 t with an acceleration of 0.5 m / s2.

k = 0.01 MN / m = 10000 N / m.

m = 2 t = 2000 kg.

a = 0.5 m / s ^ 2.

x -?

The car being towed moves with acceleration according to Newton’s 2 law: F = m * a, where F is the force that acts on it, m is the mass of the car, a is the acceleration of the car.

The car moves under the action of the cable tension force F, the value of which is determined by Hooke’s law: F = – k * x, where k is the cable stiffness, x is the absolute cable elongation.

The sign “-” means that the elastic force F is directed in the opposite direction to the elongation x.

m * a = k * x.

x = m * a / k.

x = 2000 kg * 0.5 m / s ^ 2/10000 N / m = 0.1 m.

Answer: The elongation of the towing cable is x = 0.1 m.