Question

A student uses a spring to launch a marble vertically in the air . the mass of the marble is 0.002kg and when the spring is stretched 0.05m it exerts a force of 20N. what is the maximum height the marble can reach ?

Answer 1

from the equestion:
F=mh÷g
20N=0.002×0.05÷10
let make pair subject:
hm=Fg
0.05m×0.002=20N×10
0.05m×0.002=200
0.05m=200-0.002=199.98 approximately 200.
0.05m÷0.05m=200÷0.05
m=4000
therefore; Maximum height is 4000m.