Question

A 50g toy car is spring loaded to propel forward when released. If the spring is stretched a distance of 2cm and has a K value of 500N/m. Calculate the speed at which the toy will move when the spring is released.

Answer 1

Given, spring constant in the toy car is K = 500 N / m, and, it was stretched by d = 2 cm = 0.02 m.

Mass of the car m = 50 g = 0.05 kg, and, its speed just after releasing the spring be v.

Now, according to the energy conservation principle,

Kinteic energy of the car after releasing the spring = Potential energy stored in the spring before releasing

or, mv² / 2 = Kd² / 2

or, v = $\sqrt{}$ ( Kd² / m ) = $\sqrt{}$ ( 500 x 0.02² / 0.05 ) m / s = 2 m / s.

Hence, the toy will move at 2 m / s when the spring is released.