Amass of 30.0 grams hangs at rest from the lower end of a long vertical spring. you add different amounts of additional mass δm to the end of the spring and measure the increase in length δl of the spring due to the additional mass. you plot your data in the form δl (vertical axis) versus δm (horizontal axis). plotted this way, your data lies very close to a straight line that has slope 0.0364 m/kg. what is the force constant k of the spring? use g = 9.80 m/s2.
For a spring the elastic constant is the amount of force required for the deformation of unit length in the spring. It is usually denoted by k.
We have a relation between the force, length of deformation and elastic constant given as:
= change in length
where the negative sign just denotes that the force is acting in opposite direction to the displacement.
Here we are given with the slope of Δx versus changing mass which is .
For k we need the inverse of this value multiplied by the gravity to get the corresponding load of the given mass.
According to Hooke's law:
avoiding the minus symbol and substituting the force to the gravitational force, we have:
The spring constant is 366N/m
K = 373.13 N/m
The force of the spring is equals to:
Fe - m*g = 0 => Fe = m*g
Using Hook's law:
K*X = m*g Solving for K:
K = m/X * g
In this equation, m/X is the inverse of the given slope. So, using this value we can calculate the spring's constant:
K = 10 / 0.0268 = 373.13N/m
m= 30 gm
Additional mass = ΔM
Lets take spring coefficient = k
at the equilibrium position
F= k Δl =( m + ΔM)g
Δl = change in the spring length
k Δl = ( m + ΔM)g
Δl = m g /k + g/k ΔM 1
We know that equation of line given as
y = c + S x 2
By compare equation 2 and 1
slope = s
Given that slope S= 0.052 m/kg
S =g/ k= 0.052
9.8 = 0.052 k
So the spring constant is 188.46 N/m
good for her? ? where is the question?