When a 0.101 kg mass is suspended at rest from a certain spring, the spring stretches 3.80 cm. Find the instantaneous acceleration of the mass when it is raised 5.90 cm, compressing the spring 2.10 cm.
2 answers
Yes
w=kx
.101g=k*3.80
k=.101g/3.80
force of compression=
= (.101g/.0389)(.0590-.0380)
figure that out, then
using f=ma or a=f/m
and we have to add g because the object is going down,
a=g+(.101g/.0389)(.0590-.0380) /.101g)
.101g=k*3.80
k=.101g/3.80
force of compression=
= (.101g/.0389)(.0590-.0380)
figure that out, then
using f=ma or a=f/m
and we have to add g because the object is going down,
a=g+(.101g/.0389)(.0590-.0380) /.101g)