A certain spring stretches 5.8cm when a load of 38N is suspended from it. How much will the spring stretch if 48N is suspended from it (and it doesn't reach its elastic limit)?

7.3246

The length that a spring is stretched beyond its natural position is given by L = 2W where work, W, is k

in foot-pounds and k is a constant for the given spring. If a certain spring has a constant of 66.8, and the spring is to be stretched a length of 3.2 feet beyond its natural position, how much work will be necessary?

To find out how much the spring will stretch when a load of 48N is suspended from it, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.

Hooke's Law can be expressed as:

F = k * x

Where:
F is the force applied to the spring (in Newtons)
k is the spring constant (in Newtons per meter)
x is the displacement of the spring from its equilibrium position (in meters)

To find the spring constant (k), we can rearrange the equation as:

k = F / x

Now, let's plug in the given values.

F = 38N
x = 5.8cm = 0.058m

k = 38N / 0.058m
k ≈ 655.17 N/m

Now that we have the spring constant, we can use it to calculate the new displacement (x') when a load of 48N is suspended from the spring.

F' = 48N

Using Hooke's Law:

F' = k * x'

Rearranging the equation to solve for x':

x' = F' / k

Plugging in the values:

x' = 48N / 655.17 N/m
x' ≈ 0.0733m

Therefore, the spring will stretch approximately 0.0733 meters (or 7.33 cm) when a load of 48N is suspended from it.

evidently the spring constant is

5.8cm/38N = .1526 cm/N

So, if 48N is suspended, the stretch is

48N*0.1526 cm/N = ? cm