A force of 5n stretches an elastic material by 65mm, what additional force on 1 stretch the material 70mm assumed the elastic limit is not exceeded
Using Hooke's Law: F = kx
65 mm = 0.065m, 70 mm = 0.070 m
k = 5 N/0.065 m = 76.9 N/m
F = 76.9 N/m * 0.070 m - 5 N = 0.383 N
Therefore, the additional force should be about 0.383 N.
To determine the additional force required to stretch the material by 70mm, we can use Hooke's Law, which states that the force needed to stretch or compress an elastic material is directly proportional to the extension or compression of the material, as long as the elastic limit is not exceeded.
Hooke's Law can be expressed as:
F = k * x
Where:
F is the force applied to the material,
k is the spring constant (a measure of the stiffness of the material),
x is the extension or compression of the material.
Given that a force of 5N stretches the material by 65mm, we can use this information to find the spring constant, k.
Rearranging Hooke's Law, we have:
k = F / x
Plugging in the values, we have:
k = 5N / 65mm
Now that we have the spring constant, we can determine the additional force required to stretch the material by 70mm.
Using Hooke's Law, we can rearrange the equation as follows:
F = k * x
Plugging in the values, we have:
F = (5N / 65mm) * 70mm
Calculating this, we find:
F ≈ 5.38N
Therefore, an additional force of approximately 5.38N is required to stretch the material by 70mm, assuming the elastic limit is not exceeded.
F = kx, so
70/65 the stretch means 70/65 the force
1/13 * 5 = 5/13 or 0.3846N