A load of 4N extends a spring by 10mm. What load would extend it by 15mm?
F= k x
4 Newtons = k* 10 mm
k = 0.40 N/mm
F = 0.40 N/mm * 15mm
F = 6 N
Well, well, well, looks like we've got a spring aficionado here! Alrighty then, let's unravel this conundrum. So, the load of 4N extended the spring by 10mm? That's no small feat for a spring, I must say. Now, if we want to extend it by 15mm, we'll need to give it a little extra oomph!
You know, springs can be quite sensitive. They're like Goldilocks searching for the perfect porridge—too little load and they won't extend enough, too much load and they'll extend too much! But fret not, my spring-stretching friend, I'll crunch some numbers for you.
If the load of 4N extended the spring by 10mm, we can use a little thing called proportionality to find out the load we need to stretch it by 15mm. Let's set up a ratio here: 4N is to 10mm as X is to 15mm. Cross multiply that, divide by 10, carry the one, and...
Ta-da! The load that will extend the spring by 15mm is approximately 6N. Just like Goldilocks finding that perfect bowl of porridge—just the right amount of load for our little spring friend!
Hope that brings a little "spring" to your step! If you have any more questions, I'm here to keep the laughs rollin'!
Load 1= 4N
Extension 1= 10mm
Load 2 = ?
Extension 2 = 15mm
Then,
Load=(4N×15mm)/10mm
=6N
Yes, that's another way to solve the problem using Hooke's law, which states that the force (load) applied to a spring is proportional to the extension of the spring. So if you know the constant of proportionality (spring constant), you can use it to find the force required to produce a certain extension. In this case, we have:
F1 = kx1 (where F1 = 4 N and x1 = 10 mm)
F2 = kx2 (where we need to find F2 for x2 = 15 mm)
Dividing the second equation by the first one, we get:
F2/F1 = x2/x1
Solving for F2, we get:
F2 = F1 * x2/x1 = 4 N * 15/10 = 6 N
So both methods lead to the same result, which is the force required to extend the spring by 15 mm.
To find the load that would extend the spring by 15mm, we can use Hooke's Law, which states that the force (load) applied to a spring is directly proportional to the displacement or extension of the spring. Hooke's Law can be mathematically represented as:
F = k * x
Where:
F is the force (load) applied to the spring
k is the spring constant
x is the displacement or extension of the spring
Given that the load of 4N extends the spring by 10mm, we can use this information to determine the spring constant (k). Rearranging the equation, we have:
k = F / x
k = 4N / 10mm
k = 0.4 N/mm
Now, we can use the spring constant to find the load required to extend the spring by 15mm. Rearranging the equation one more time:
F = k * x
F = 0.4 N/mm * 15mm
F = 6 N
Therefore, the load required to extend the spring by 15mm is 6N.