An elastic string of length 20cm extends to 24cm. When it supports a weight of 50N, the energy stored in the string is
= (24m - 20m) x 50N
= 4 x 50
= 200J
To find the energy stored in an elastic string, we can use the formula:
Potential Energy (U) = (1/2) * k * x^2
where,
k is the spring constant
x is the extension or displacement of the string from its equilibrium position
In this case, the length of the string extends from 20cm to 24cm, which means the extension (x) is 4cm or 0.04m (since 1cm = 0.01m).
First, let's calculate the spring constant (k):
k = F / x
where,
F is the force applied (given as 50N)
x is the extension (0.04m)
k = 50N / 0.04m
k = 1250 N/m
Now, with the spring constant and extension, we can calculate the potential energy (U):
U = (1/2) * k * x^2
U = (1/2) * 1250 N/m * (0.04m)^2
U = (1/2) * 1250 N/m * 0.0016 m^2
U = 1 N * m
Therefore, the energy stored in the string is 1 Joule (J).
To calculate the energy stored in the elastic string, we can use Hooke's Law, which states that the force exerted by an elastic string is directly proportional to the extension of the string.
The equation for Hooke's Law is:
F = k * x
Where:
F = Force applied to the string (in Newtons)
k = Spring constant (a measure of the stiffness of the string)
x = Extension of the string (the difference between the final length and the original length)
Given that the original length of the string is 20cm and it extends to 24cm, the extension (x) would be:
x = Final length - Original length
x = 24cm - 20cm
x = 4cm
Now, let's find the spring constant (k). Since we are given the force (F) as 50N, we can rearrange the Hooke's Law equation to solve for k:
k = F / x
k = 50N / 4cm
Before proceeding with the calculation, we need to convert centimeters (cm) to meters (m) since the unit of force is Newton (N), and the SI unit of length is meter (m):
1m = 100cm
Therefore, 4cm is equal to 4/100 m, or 0.04m.
Now we can calculate the spring constant (k):
k = 50N / 0.04m
k = 1250 N/m
Now that we have the spring constant, we can calculate the energy stored in the string using the formula:
Energy = (1/2) * k * x^2
Plugging in the values:
Energy = (1/2) * 1250 N/m * (0.04m)^2
Energy = 0.5 * 1250 N/m * 0.0016 m^2
Energy = 1.25 J
Therefore, the energy stored in the elastic string when it supports a weight of 50N is 1.25 Joules.
(.24 m - .20 m) * 50 N = ? Joules
Well, that's quite a stretch! The energy stored in the string can be calculated using Hooke's Law. The formula is:
E = 1/2 * k * x^2
Where E is the energy stored, k is the spring constant, and x is the extension of the string.
Now, we are given that the string extends from 20cm to 24cm, so the extension is 4cm (or 0.04m). However, we don't know the spring constant of the string (k) in this case, so it's hard to calculate the exact energy stored.
So, I'm afraid I can't give you an answer without that information. But hey, at least we stretched our minds a bit with this one!