Question

A 2.0-kg mass is attached to one end of a spring with a spring constant of 100 N/m and a 4.0-kg mass is attached to the other end. The masses are placed on a horizontal frictionless surface and the spring is compressed 10 cm. The spring is then released with the masses at rest and the masses oscillate. When the spring has its equilibrium length for the first time the 2.0-kg mass has a speed of 0.36 m/s. The mechanical energy that has been lost to this instant is:

Answer 1

mechanical energy that has been lost to this instant is 0.3056 J

Step-by-step explanation:

given data

mass m1 = 2 kg

spring constant k = 100 N/m

mass m2 = 4 kg

spring compressed x = 10 cm

speed = 0.36 m/s

solution

we know that initial energy is

initial energy= 0.5 k x²

initial energy = 0.5 × 100 × 0.1²

initial energy = 0.5 J

and

momentum conservation is

2 × 0.36 = 4 v

so v = 0.18 m/s

and

energy lost will be here

energy lost = 0.5 - (0.5 × 2 × 0.36² + 0.5 × 4 × 0.18²)

energy lost = 0.3056 J

so mechanical energy that has been lost to this instant is 0.3056 J