Answer:
Both springs have a constant of 25Nm and the block is motionless. If the bottom spring is compressed 0.4m past its equilibrium and the block has a mass of 3kg, how far is the top spring stretched past its equilibrium?
g=10ms2
Possible Answers:
1.0m
0.8m
0.2m
0.4m
0.6m
Correct answer:
0.8m
Step-by-step explanation:
Since the block is motionless, we know that our forces will cancel out:
Fnet=0
There are three forces in play: one from each spring, as well as the force of gravity. If we assume that forces pointing up are positive, we can write:
Fspring,top+Fspring,bot−mg=0
Plugging in expressions for each spring force, we get:
kxtop+kxbot−mg=0
Rearring for the displacement of the top spring, we get:
xtop=mg−kxbotk=(3kg)(10ms2)−(25Nm)(0.4m)25Nm
xtop=30N−10N25Nm=0.8m