Answer:
![x=0.0049\ m= 4.9\ mm](https://img.qammunity.org/2020/formulas/physics/high-school/fqgcq8ezslls7ywhnfzord9wx294svkzop.png)
![d=0.01153\ m=11.53\ mm](https://img.qammunity.org/2020/formulas/physics/high-school/sm73rp31mkg9gr3f88901738s7y97v429v.png)
Step-by-step explanation:
Given:
- mass of the object,
![m=0.75\ kg](https://img.qammunity.org/2020/formulas/physics/high-school/shoa06lv3bjcezsiu8919ithzmqbp2wgab.png)
- elastic constant of the connected spring,
![k=150\ N.m^(-1)](https://img.qammunity.org/2020/formulas/physics/high-school/60kmgq7dltixt9kj7ydvt42dnsrm9qet9t.png)
- coefficient of static friction between the object and the surface,
![\mu_s=0.1](https://img.qammunity.org/2020/formulas/physics/college/2hf1ydytdpag08m4gjgkesahjl37buxns6.png)
(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.
![f=k.x](https://img.qammunity.org/2020/formulas/physics/high-school/2ye9m70lcdn0vi99hywki888hz1inj809z.png)
![\mu_s.N=k.x](https://img.qammunity.org/2020/formulas/physics/high-school/oqrmc6z5d8w7xtqx14keta9te5q1412yuj.png)
where:
N = m.g = the normal reaction force acting on the body under steady state.
![0.1* (9.8* 0.75)=150* x](https://img.qammunity.org/2020/formulas/physics/high-school/713uaxwh52ik6ln480fx02b7sos9zix9mb.png)
![x=0.0049\ m= 4.9\ mm](https://img.qammunity.org/2020/formulas/physics/high-school/fqgcq8ezslls7ywhnfzord9wx294svkzop.png)
(b)
Now, according to the question:
- Amplitude of oscillation,
![A= 0.0098\ m](https://img.qammunity.org/2020/formulas/physics/high-school/ceyf9zcee3kjggo13falqq07hevzkoyetk.png)
- coefficient of kinetic friction between the object and the surface,
![\mu_k=0.085](https://img.qammunity.org/2020/formulas/physics/high-school/gp1776gpq6r88makyhzjy9impmswshfmql.png)
Let d be the total distance the object travels before stopping.
Now, the energy stored in the spring due to vibration of amplitude:
![U=(1)/(2) k.A^2](https://img.qammunity.org/2020/formulas/physics/high-school/qy3h6o9m6gh3o27uuf0fg3dbcf8a3dgvbs.png)
This energy will be equal to the work done by the kinetic friction to stop it.
![U=F_k.d](https://img.qammunity.org/2020/formulas/physics/high-school/cl0pgb34r3hahexjqub1t40hd3tkeqfx0u.png)
![(1)/(2) k.A^2=\mu_k.N.d](https://img.qammunity.org/2020/formulas/physics/high-school/gzffh0lw5xe0prqls5kvymspx2drvfspxa.png)
![0.5* 150* 0.0098^2=0.0850 * 0.75* 9.8* d](https://img.qammunity.org/2020/formulas/physics/high-school/m4tbyi9gkynrfz8oofblaqwkbwb3srpu80.png)
![d=0.01153\ m=11.53\ mm](https://img.qammunity.org/2020/formulas/physics/high-school/sm73rp31mkg9gr3f88901738s7y97v429v.png)
is the total distance does it travel before stopping.