A block is attached to the top of a spring that stands vertically on a table. The spring stiffness is 64 N/m, its relaxed length is 31 cm, and the mass of the block is 225 g. The block is oscillating up and down as the spring stretches and compresses. At a particular time you observe that the velocity of the block is <0, 0.0877, 0> m/s and the position of the block is <0, 0.0798, 0> m relative to an origin at the base of the spring. Using a time step of 0.1 s, determine the position of the block 0.2 s later.
answer) force on the block is
F=mg=0.225kg*9,8m/s2=2.205N
Force on the block due to spring
f=-kx=-64*0.31=-19.84N
so net force acting on the block
Fnet=2.205-19.84=-17.635N
from the law of conservation of momentum
P=Fnet*t
Pf-Pi=Fnet*t
Pf-Pi=-17.635N*0.1s=-1.7635
pi=mvi=0.225*0.0877=0.0197
Pf=-1.7635+0.0197=-1.7438
using the relation
Vavg=Pf/m
Vavg=x2-x11/t
x2-x1/t=Pf/m
x2-0.0798/0.2=-1.7348/0.225
x2=1.4622m
so the answer is -1.4622m or
<0,-1.4622,0>
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