Answer :
The total distance traveled by the block before it comes to rest is 3.517 ft.
Explanation :
Given that,
Weight of block = 25 lb
Initial speed = 10 ft/s
Coefficient of kinetic friction = 0.4
Spring constant [tex]k_{A}=10\ lb/in[/tex]
Spring constant [tex]k_{B}= 60\ lb/in[/tex]
The kinetic energy of the system
[tex]K.E_{1}=\dfrac{1}{2}mv_{0}^2[/tex]
[tex]K.E_{1}=\dfrac{1}{2}\times(\dfrac{W}{g})v_{0}^2[/tex]
We need to calculate the frictional force
Using formula of frictional force
[tex]F=\mu\times W[/tex]
Put the value into the formula
[tex]F=0.4\times25[/tex]
[tex]F=10\ N[/tex]
The potential work due to spring is zero because the block comes to rest.
We need to calculate the distance
Using work energy theorem
[tex]K.E_{1}+\sum U=T_{2}[/tex]
[tex]K.E_{1}-W_{f}-W_{s}=\dfrac{1}{2}k_{b}s^2[/tex]
Put the value in the equation
[tex]\dfrac{1}{2}\times(\dfrac{W}{g})v_{0}^2-10\times (1+s_{1})-\dfrac{1}{2}k_{b}s_{3}^2=0[/tex]
Here, [tex]s_{1}=s_{3}[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times(\dfrac{25}{32.2})\times10^2-10+10s-\dfrac{1}{2}\times60\times s^2=0[/tex]
[tex]38.81-10+10s_{1}-30s_{1}^2=0[/tex]
[tex]28.81+10s-30s^2=0[/tex]
[tex]s_{1}=0.827\ ft[/tex]
We need to calculate the distance s₂
Using work energy theorem
[tex]K.E_{2}+\sum U=T_{3}[/tex]
[tex]0+\dfrac{1}{2}k_{b}s_{3}^2-k_{A}(s_{1}+s_{2})=0[/tex]
Put the value in to the formula
[tex]\dfrac{1}{2}\times60\times(0.827)^3-10(0.827+s_{2})=0[/tex]
[tex]16.9-8.27-10s_{2}=0[/tex]
[tex]8.63-10s_{2}=0[/tex]
[tex]s_{2}=\dfrac{8.63}{10}[/tex]
[tex]s_{2}=0.863\ ft[/tex]
We need to calculate the total distance
[tex]s=2s_{1}+s_{2}+1[/tex]
Put the value in the equation
[tex]s=2\times0.827+0.863+1[/tex]
[tex]s=3.517\ ft[/tex]
Hence, The total distance traveled by the block before it comes to rest is 3.517 ft.
