In the parallel spring system, the springs
are positioned so that the 44 N weight
stretches each spring equally. The spring con-
stant for the left-hand spring is 3.6 N/cm and
the spring constant for the righ-hand spring
is 6.4 N/cm.
How far down will the 44 N weight stretch
the springs?
Answer in units of cm.
The forces are additive.
F1+F2=44N
3.6*x+6.4x=44
solve for x.
To find out how far down the 44 N weight will stretch the springs, we can use Hooke's Law, which states that the force applied to a spring is directly proportional to the displacement it causes.
Hooke's Law can be expressed as: F = kx
Where:
F is the force applied to the spring
k is the spring constant
x is the displacement of the spring
Since the weight is stretching both springs equally, the force applied to each spring is the same. Therefore, we can set up two equations using Hooke's Law:
F_left = k_left * x
F_right = k_right * x
Given that the weight of 44 N is applied, we have:
F_left = F_right = 44 N
Substituting the values:
3.6 * x = 44
6.4 * x = 44
Now we can solve for x.
Starting with the equation 3.6 * x = 44:
Dividing both sides by 3.6:
x = 44 / 3.6 = 12.22 cm
Now solving for x using the equation 6.4 * x = 44:
Dividing both sides by 6.4:
x = 44 / 6.4 = 6.87 cm
Since both springs are stretched equally, we can take the average of the two values of x:
(x_left + x_right) / 2 = (12.22 + 6.87) / 2 = 9.545 cm
Therefore, the 44 N weight will stretch the springs approximately 9.545 cm down.