Answer:
Explanation:
Find attached the figure to solve this problem (taken from a problem, in the internet, with the same statement, but different mass for the blok).
The block will stop when all its kinetic energy is absorbed by the friction and the spring.
1. Initial kinetic energy of the blockm [tex]KE_i[/tex]
[tex]KE_i=\dfrac{1}{2}mass\times (speed)^2\\\\\\KE_i=\dfrac{1}{2}(5.6kg)\times (5m/s)^2=70J[/tex]
2. Work of friction
The friction force is the product of the normal force by the coefficient of kinetic friction , [tex]\mu_k=0.25[/tex] .
Since, the only vertical force is the force of gravity, the normal force, [tex]F_N[/tex] , is the weight of the block:
[tex]F_N=5.6kg\times 9.8m/s^2=54.88N[/tex]
Then, the friction force, [tex]F_f[/tex] , is:
[tex]F_f=0.25\times 54.88N=13.72N[/tex]
The distance run by the block before stopping is the 2 meters distance plus the amount the spring compresses. Calling x the distance the spring compresses, the friction work is:
[tex]W_f=13.72N\times (2+x)[/tex]
3. Energy absorbed by the spring
The energy absorbed by the spring is the elastic potential energy, PE, which is given by the formula:
[tex]PE=\dfrac{1}{2}kx^2[/tex]
Where k is the elasticity constant of the spring (200B/m, according to the figure), and x is the distance the spring compresses.
Substituting:
[tex]PE=\dfrac{1}{2}\times 200N/m\times x^2\\\\\\PE=100N/m\cdot x^2[/tex]
4. Final equation
Now you can write your equation to find the compression of the spring, x:
[tex]70=13.72(2+x)+100x^2[/tex]
Solving:
[tex]70=27.44+13.72x+100x^2\\\\100x^2+13.72x-42.56[/tex]
Use the quadratic formula:
[tex]x=\dfrac{-13.72\pm \sqrt{(13.72)^2-4(100)(-42.56)}}{2(100)}[/tex]
There is one negative solution, which you discard, and the positive solution is 0.59.
