# a 0.19 kg ball is placed on a compressed spring on the floor. the spring exerts an average force of 2.2 N?

a 0.19 kg ball is placed on a compressed spring on the floor. the spring exerts an average force of 2.2 N through a distance of 17 cm as it shoots the ball upward. how high will the ball travel above the released spring?

a. 0.2 m

b. 4.5 m

c. 20 m

d. 197 m

### 2 Answers

- Anonymous1 decade agoBest Answer
Use the law of conservation of energy in this problem, i.e.,

Energy from the spring = Potential energy of ball

Fx = mgh

where

F = force on spring = 2.2 N (given)

x = compression distance of spring = 17 cm = 0.17 m (given)

m = mass of ball = 0.19 kg (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

h = height at which ball will rise = unknown

Substituting values,

2.2(.17) = (0.19)(9.8)h

Solving for "h",

h = 2.2(.17)/(0.19)(9.8)

h = 0.20 meter (option "a" from your choices)

- tsistinasLv 43 years ago
The paintings completed on the ball = Fx would be equivalent to the aptitude potential of the ball while it reaches its maximum top h Fx = mgh 2.8*.15=9.80 one*.18*h h=(2.8*.15)/(9.80 one*.18)=.24m