Amass of 2.82 kg is hung from a spring, causing the spring to stretch 0.331 m. if a second mass of 3.09 kg is now added to the spring, how far with it now stretch with both masses hanging? ? state your answer to the correct number of significant digits and include the proper units.

0.694 m

Explanation:

Case 1 : When only mass of 2.82 kg is hanged from spring

m = mass hanged from the spring = 2.82 kg

x = stretch caused in the spring = 0.331 m

k = spring constant

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

k (0.331) = (2.82) (9.8)

k = 83.5 N/m

Case 2 : When both masses are hanged from spring

m = mass hanged from the spring = 3.09 + 2.82 = 5.91 kg

x = stretch caused in the spring = ?

k = spring constant = 83.5 N/m

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

(83.5) x = (5.91) (9.8)

x = 0.694 m

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s