# A 0.18kg ball is placed on a compressed spring on the floor. The spring exerts an average force of 2.8N?

through a distance of 15cm as it shoots the ball upward. How high will the ball travel above the release spring.

The answer is 0.24m. Can somebody please explain how to get this answer?

### 3 Answers

- Anonymous1 decade agoBest Answer
Use the law of conservation of energy in solving this problem.

Energy from spring = Energy absorbed by ball

Energy by spring = Fx

Energy absorbed by ball = mgh

where

F = force exerted by the spring = 2.8 N (given)

x = compressed distance of the spring = 15 cm = 0.15 m (given)

m = mass of the ball = 0.18 kg (given)

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

h = maximum height that ball will attain after being released from the spring

Substituting appropriate values,

(2.8)(0.15) = (0.18)(9.8)h

Solving for "h",

h = (2.8)(0.15)/(0.18)(9.8)

h = 0.24 m.

- oldschoolLv 71 decade ago
The work done on the ball = Fx will be equal to the potential energy of the ball when it reaches its maximum height h

Fx = mgh

2.8*.15=9.81*.18*h

h=(2.8*.15)/(9.81*.18)=.24m

- rozaLv 43 years ago
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