Answer:
k=188.46 N/m
Explanation:
Given that
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
S= g/k
Given that slope S= 0.052 m/kg
S =g/ k= 0.052
g=9.8 m/s²
9.8 = 0.052 k
k=188.46 N/m
So the spring constant is 188.46 N/m
Total work energy on the input side is WE = Fs; where F is a force acting on a mass to push it s distance. This is the so-called work function. Let fs = we, which is the work energy (useful energy) attained as output when WE is input.
<span>From the conservation of energy WE = Fs = fs - kNs = Total Output energy. Net force f = F - kN where kN is friction force acting against the pushing (input) force F. In the real world, there is always friction at some level. That is kN > 0 always. </span>
<span>Thus Fs = (F - kN)s; kNs = the energy lost to friction where k is the friction coefficient and N is the normal force on the surface(s) where the friction is generated. By definition, efficiency = fs/Fs = useful work/work input. Clearly fs = Fs - kN < Fs . Thus efficiency = fs/Fs < 1.00, which means output fs < Fs the input whenever kN > 0, which in the real world it always is. </span>
<span>The short answer is...output is always less than input because of friction and, sometimes, other losses like wind drag (which is a form of friction anyway).</span>