Respuesta :
Answer:
[tex]x=0.0049\ m= 4.9\ mm[/tex]
[tex]d=0.01153\ m=11.53\ mm[/tex]
Explanation:
Given:
- mass of the object, [tex]m=0.75\ kg[/tex]
- elastic constant of the connected spring, [tex]k=150\ N.m^{-1}[/tex]
- coefficient of static friction between the object and the surface, [tex]\mu_s=0.1[/tex]
(a)
Let x be the maximum distance of stretch without moving the mass.
The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.
[tex]f=k.x[/tex]
[tex]\mu_s.N=k.x[/tex]
where:
N = m.g = the normal reaction force acting on the body under steady state.
[tex]0.1\times (9.8\times 0.75)=150\times x[/tex]
[tex]x=0.0049\ m= 4.9\ mm[/tex]
(b)
Now, according to the question:
- Amplitude of oscillation, [tex]A= 0.0098\ m[/tex]
- coefficient of kinetic friction between the object and the surface, [tex]\mu_k=0.085[/tex]
Let d be the total distance the object travels before stopping.
Now, the energy stored in the spring due to vibration of amplitude:
[tex]U=\frac{1}{2} k.A^2[/tex]
This energy will be equal to the work done by the kinetic friction to stop it.
[tex]U=F_k.d[/tex]
[tex]\frac{1}{2} k.A^2=\mu_k.N.d[/tex]
[tex]0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d[/tex]
[tex]d=0.01153\ m=11.53\ mm[/tex]
is the total distance does it travel before stopping.
