Question

A 5.22×104 kg railroad car moves on frictionless horizontal rails until it hits a horizontal spring stopper with a force constant of 4.58×105 N/m . When the railroad car comes to a complete stop, the compression of the spring stopper is 32 cm . How fast was the railroad car initially?

Answer 1

To solve this problem we will apply the principles of conservation of energy, for which we have to preserve the initial kinetic energy as elastic potential energy at the end of the movement. If said equality is maintained then we can affirm that,

$\text{Initial Energy}=\text{Final Energy}$

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

Here,

m = mass

k = Spring constant

x = Displacement

v = Velocity

Rearranging to find the velocity,

$mv^2 = kx^2$

$v^2 = \frac{kx^2}{m}$

$v = \sqrt{\frac{kx^2}{m}}$

Our values are,

$m = 5.22*10^4kg$

$k = 4.58*10^5N/m$

$x = 32cm = 0.32m$

Replacing our values we have,

$v = \sqrt{\frac{(4.58*10^5)(5.22*10^4)}{0.32}}$

$v = 2.733*10^5m/s$

Therefore the velocity is $2.733*10^5m/s$