A certain spring stretches 6 cm when a load of 30 N is suspended from it. How much will the spring stretch if 50 N is suspended from it (and it doesn't rsach its elastic limit)? I don't understand what to do for this problem.
deformation is directly proportional to distance/
50/30=x/6
x=10 cm
Ah, the magic of springs! Let me break it down for you in a way that will hopefully make you crack a smile.
So, we know that when a load of 30 N is suspended from the spring, it stretches 6 cm. Now, imagine this spring as a perfect little trampoline. When you bounce on it with extra weight, it stretches even more.
To find out how much the spring will stretch when 50 N is suspended from it, we can set up a proportion. Kind of like when you're trying to decide how many cookies to eat, but don't want to overdo it.
Here's the formula:
Load 1 / Stretch 1 = Load 2 / Stretch 2
In this case:
30 N / 6 cm = 50 N / x cm
Cross-multiplying, we get:
30 N * x cm = 50 N * 6 cm
Now, it's time for some simple math:
30x = 300
Divide both sides by 30 to get:
x = 10
So, the spring will stretch approximately 10 cm when 50 N is suspended from it. Who knew stretching springs could be such a fun math exercise? Spring on!
To solve this problem, we need to use Hooke's Law, which 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 written as:
F = k * x
Where:
F is the force exerted by the spring (in Newtons),
k is the spring constant (in Newtons per meter), and
x is the displacement of the spring from its equilibrium position (in meters).
In this case, we are given that a load of 30 N stretches the spring by 6 cm. To find the spring constant, we can rearrange Hooke's Law equation:
k = F / x
Plugging in the values, we have:
k = 30 N / 0.06 m
k = 500 N/m
Now, we can use this spring constant to find the stretch of the spring when a 50 N load is suspended from it.
F = 50 N
k = 500 N/m
Using Hooke's Law equation:
F = k * x
Rearranging the equation to solve for x:
x = F / k
Plugging in the values:
x = 50 N / 500 N/m
x = 0.1 m or 10 cm
Therefore, the spring will stretch by 10 cm when a 50 N load is suspended from it.
To solve this problem, you need to understand Hooke's Law, which states that the force required to stretch or compress a spring is directly proportional to the displacement from its equilibrium position.
In mathematical terms, Hooke's Law can be expressed as F = k * x, where F is the force applied to the spring, k is the spring constant, and x is the displacement from the equilibrium position.
To find the spring constant, you need to use the given information: the force and the displacement. In this case, the force is 30 N, and the displacement is 6 cm (which can be converted to meters by dividing by 100, giving you 0.06 m).
Now, rearrange Hooke's Law equation to solve for k: k = F / x. Plugging in the values, you get k = 30 N / 0.06 m, which simplifies to k = 500 N/m.
Once you have the spring constant, you can use it to calculate the spring's stretch when a different force is applied. In this case, the force is 50 N.
Using the formula F = k * x and rearranging it to solve for x, you can calculate the stretch:
x = F / k = 50 N / 500 N/m = 0.1 m.
To convert this to centimeters, multiply by 100, giving you x = 10 cm.
Therefore, the spring will stretch 10 cm if a 50 N load is suspended from it, assuming it does not reach its elastic limit.