When a 4kg mass is hung vertically on a light spring, the spring stretches 2.5cm. a.) how far will the spring stretch if an additional 1.5kg mass is hung on it? b.) if the 4 kg mass is removed, how far will the spring stretch if a 1.5 kg mass is hung on it? c.) how much work must be done on the spring to stretch the same spring 4 cm from its equilibrium position?

Explanation:

The weight of the 4kg mass equals the force of the spring.

mg = kx

(4 kg) (9.8 m/s²) = k (0.025 m)

k = 1568 N/m

If an additional 1.5 kg is added, the spring stretches to:

(4 kg + 1.5 kg) (9.8 m/s²) = (1568 N/m) x

x = 0.034 m

x = 3.4 cm

If the 4 kg is removed, the spring stretches to:

(1.5 kg) (9.8 m/s²) = (1568 N/m) x

x = 0.0094 m

x = 0.94 cm

The work done to stretch a spring a distance x is:

E = 1/2 k x²

E = 1/2 (1568 N/m) (0.04 m)²

E = 1.3 J

found in the nucleus

has mass of one amu