Potential Energy (WebAssign Question)

[queenspublic]

Homework Statement

A 4.8 g marble is fired vertically upward using a spring gun. The spring must be compressed 8.0 cm if the marble is to just reach a target 27 m above the marble's position on the compressed spring.

(a) What is the change in the gravitational potential energy of the marble-Earth system during the 27 m ascent?
(a) ? J

(b) What is the change in the elastic potential energy of the spring during its launch of the marble?
(b) ? J

(c) What is the spring constant of the spring?
(c) ? N/m

2. The attempt at a solution

(a)
27 - .080 = 26.92
(26.92)(9.8)(.0048) = (1/2)(.0048)v^2
v = 22.97 <<<< Isn't this the answer? But it's wrong.

(b)
(1/2)(the answer to c: ?)(.080)^2

(c)
(1/2)(.0048)(the answer to a: 22.97?)^2 + (.0048)(9.8)(.080) = (1/2)(k)(.080)^2
k = ?

 

[rl.bhat]

a) mgh
b) 1/2*k*x^2
From a) and b) you can calculate c)

 

[The Liberator]

What do you state gravity as?

the gravitational potential energy equation is this:

$$U_g=mgh$$

"U" is the energy in joules, "m" is mass in kg, "g" is the gravity value in ms-², and "h" is the maximum height of the object in metres. the value of "g" is normally taken as 9.8 or 10ms-1².

That might clear some of the errors up. :)