Question

Majin buu makes his lower body in to a spring, to bounce off the wall to fly in the air 600 m to punch Gogeta. If Majin buu is 45 kg and the spring compressed 5 m what is the spring's k constant.

Answer 1

Take into account that the elastic potential energy gained when Majin compressed the spring must be equal to the gravitational potential energy gained by Majin when he is at 600m over the ground (this is mandatory by the law of conservation of energy).

Then, you can write:

`(1)/(2)kx^2=\text{mgh}`

The left hand side of the equation represents the elastic potential energy and the right hand side the gravitational potential energy.

k: spring constant = ?

g: gravitational acceleration constant = 9.8m/s^2

h: height = 600 m

x: compression of the spring = 5m

m: mass of Majin = 45 kg

Solve the equation above for k, replace the values of the other parameters and simplify:

`\begin{gathered} k=\frac{2\text{mgh}}{x^2} \\ k=(2(45kg)(9.8(m)/(s^2))(600m))/((5m)^2) \\ k=21168(N)/(m) \end{gathered}`

Hence, the spring's constant is 21168N/m