The length of a spring when a mass of 20g is hung from its end is 14cm,while its total length is 16cm when a mass of 30g is hung from the same end.calculate the unstretched spring assuming Hooke's law is obeyed.
k = (M2-M1)/(L2-L1) = (30-20)/(16-14) = 5 g/cm.
d = 20g * 1cm/5g = 4cm = Distance stretched.
Lu = 14-4 = 10cm = Un-stretched length.
Well we are all learning, but I personally don't understand how you get that Distance Stretched and the Unstreched Lenght please........
If you can do so I will be so grateful.
Ope I want to understand how u got that 5 that you said 20/5. How did u get it 🤕😵
I understand the unstretch3d length but where did u get 1 feom
I understand the unstretch3d length but where did u get 1 from
To calculate the unstretched length of the spring, we need to use Hooke's law and the provided information.
Hooke's law states that the force applied to a spring is directly proportional to the extension or compression of the spring as long as the spring remains within its elastic limit. Mathematically, Hooke's law is represented as:
F = k * x
Where:
F is the force applied to the spring
k is the spring constant (a measure of the stiffness of the spring)
x is the extension or compression of the spring
In this case, we can assume that the spring is stretched, so the extension of the spring (x) can be calculated as the difference between the final length of the spring (L) and its original length (Lo). Therefore, we have:
x = L - Lo
Let's calculate the spring constant (k) first. We can do so by using the information provided for both masses.
For the 20g mass:
F1 = m1 * g
F1 = 0.02 kg * 9.8 m/s^2
F1 = 0.196 N
For the 30g mass:
F2 = m2 * g
F2 = 0.03 kg * 9.8 m/s^2
F2 = 0.294 N
Using Hooke's law, we can set up the following equations:
F1 = k * x1
F2 = k * x2
Rearranging the equations, we get:
k = F1 / x1
k = F2 / x2
Substituting the values:
k = 0.196 N / x1
k = 0.294 N / x2
Now, we can equate the two expressions for k:
0.196 N / x1 = 0.294 N / x2
Cross-multiplying:
0.196 N * x2 = 0.294 N * x1
Simplifying further:
x2 / x1 = 0.294 N / 0.196 N
x2 / x1 = 1.5
Since we know that the extension when the 20g mass is hung is 14 cm and the extension when the 30g mass is hung is 16 cm, we can express this as:
x1 = 0.14 m
x2 = 0.16 m
Substituting the values into the equation:
0.16 m / 0.14 m = 1.5
Simplifying:
1.1428571428571428 ≈ 1.5
From this, we can conclude that x2 is approximately 1.5 times larger than x1.
To find the unstretched length of the spring (Lo), we need to calculate x when F = 0. Using Hooke's law:
F = k * x
0 = k * x
Since the force applied is 0 N, we can set up the following equation:
0 = k * x
Solving for x gives us the extension at the unstretched length.
0 = k * x
0 = k * Lo
Therefore, Lo = 0
This means that the unstretched length of the spring is 0 cm.
For the second step F= ke
Therefore, e=F/k = 20/5 =4cm
For the unstreched length (original length) = L1 - e = 14 - 4 = 10cm