Now consider the same two springs in series.
6.4 N/cm, 3.6 N/cm
44 N
What distance will the spring of constant
3.6 N/cm stretch?
Answer in units of cm.
In this same series spring system, what is the
effective combined spring constant kseries of
the two springs?
Answer in units of N/cm.
The force is the same in series.
44=F1=k1*x1
44=k2*x2
so total displacement is x1+x2
x1+x2=total= 44(1/k1+1/k2)
To find the distance that the spring of constant 3.6 N/cm will stretch, we need to consider the total force exerted on the combined springs.
First, let's calculate the effective spring constant of the two springs in series. When springs are connected in series, the effective spring constant is given by the equation:
1/keff = 1/k1 + 1/k2
where keff is the effective spring constant and k1, k2 are the individual spring constants.
Given:
k1 = 6.4 N/cm
k2 = 3.6 N/cm
Plugging these values into the equation, we can solve for keff:
1/keff = 1/6.4 N/cm + 1/3.6 N/cm
1/keff = (3.6 + 6.4) / (6.4 * 3.6)
1/keff = 10 / 23.04
1/keff = 0.4348
Now, we can find the total force exerted on the combined springs by multiplying the effective spring constant by the distance stretched:
Force = keff * distance
Given:
Force = 44 N
keff = 0.4348 N/cm
Rearranging the equation, we can solve for the distance:
distance = Force / keff
distance = 44 N / 0.4348 N/cm
Now, let's calculate the distance:
distance = 101.03 cm
Therefore, the spring of constant 3.6 N/cm will stretch approximately 101.03 cm.