An elastic chord can be stretched to its elastic limit by a load of2N. If a 35cm length of the chord is extended 0.6 cm by a force of 0.5N what will be the length of the chord when the stretching force is 2.5N
Let the original length of the chord be L.
Since a load of 2N stretches the chord to its elastic limit, we can write:
2N = kL
where k is the spring constant of the chord.
To find k, we can use the information given in the second part of the problem:
When a force of 0.5N is applied, the chord is extended by 0.6 cm. So we have:
0.5N = k(35cm + 0.6cm)
Simplifying this, we get:
k = 0.5N / 35.6cm = 0.014 cm^-1
Now, let's find the new length of the chord when a stretching force of 2.5N is applied.
We can use Hooke's law:
F = kx
where F is the stretching force, x is the extension of the chord, and k is the spring constant we just found.
We want to find the new length L' when the force is 2.5N and the extension is x'. So we have:
2.5N = k(L' - L)
0.5cm = x' - 0.6cm (since the original length is 35cm)
Now we can solve for L':
L' = (2.5N + 2N)/k + L
L' = (4.5N) / 0.014 cm^-1 + 35cm
L' = 364.29 cm
Therefore, the length of the chord when the stretching force is 2.5N will be 364.29 cm.
A spring of force constant 1500Nm is acted upon by a constant force of 75N calculate the potential energy stored in the spring
The potential energy stored in a spring is given by the formula:
U = (1/2)kx^2
where U is the potential energy, k is the force constant of the spring, and x is the displacement of the spring from its equilibrium position.
In this problem, we are given that the force constant of the spring is 1500 N/m. We are also told that a constant force of 75 N is acting on the spring. In order to calculate the potential energy stored in the spring, we need to determine how much the spring is displaced from its equilibrium position as a result of this applied force.
We know that the force acting on the spring is equal to the force constant times the displacement of the spring:
F = kx
In this case, the force acting on the spring is 75 N, and the force constant is 1500 N/m. Solving for x, we get:
x = F/k = 75 N / 1500 N/m = 0.05 m
So the spring is displaced by 0.05 meters as a result of the applied force.
Now we can calculate the potential energy stored in the spring:
U = (1/2)kx^2 = (1/2)(1500 N/m)(0.05 m)^2 = 2.25 J
Therefore, the potential energy stored in the spring is 2.25 joules.
To solve this problem, we need to use Hooke's Law, which states that the force required to extend or compress a spring is directly proportional to the displacement (change in length) produced.
First, let's calculate the spring constant (k) using the given information. We know that a load of 2N stretches the chord to its elastic limit, and a 35cm length is extended by 0.6cm with a force of 0.5N. We can set up the equation as follows:
k = F / x
where k is the spring constant, F is the force applied, and x is the change in length.
k = 0.5N / 0.6cm = 0.833 N/cm
Now, we can use this spring constant to calculate the change in length when the stretching force is 2.5N. Let's call this change in length y. We can set up another equation:
2.5N = k * y
Substituting the value of k, we get:
2.5N = 0.833 N/cm * y
Now, solve for y:
y = 2.5N / 0.833 N/cm ≈ 3 cm
Therefore, the length of the chord when the stretching force is 2.5N will be:
35cm + 3cm = 38 cm.