Question

A 1300 kg car rolling on a horizontal surface has speed v = 56 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.5 m . Part A What is the spring constant of the spring? Express your answer using two significant figures. IVO AED OD? N/m

Answer 1

Given the mass of the car m = 1300kg. the velocity of the car = 56km/h = 15.55m/s and the distance compresses by the spring is x = 2.5m.

Here the kinetic energy of the car is converted to the potential energy of the spring. therefore

$\frac{1}{2}mv^{2}=\frac{1}{2}kx^{2}$

$mv^{2}=kx^{2}$

$k=\frac{mv^{2}}{x^{2}}=\frac{1300\times (15.55)^{2}}{(2.5)^{2}}=50294.92N/m$

The spring constant of the spring is 50294.92N/m.