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 â 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' â 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