Answer:
Final velocity will be 7.31 m/sec
Explanation:
We have given applied force F = 150 N
Mass of the object m = 11 KG
Distance traveled by the object s = 2.5 m
Coefficient of kinetic friction [tex]\mu _k=0.3[/tex]
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
So frictional force applied on the object [tex]f=-\mu _kmg=-0.3\times 11\times 9.8=-32.34N[/tex]
So net force = 150-32.34 = 117.66 N
As work done us equal to [tex]W=Fs[/tex]
So work done = [tex]117.66\times 2.5=294.15j[/tex]
Now according to work energy theorem work done is equal to change in kinetic energy
So [tex]\frac{1}{2}mv^2=294.15[/tex]
[tex]\frac{1}{2}\times 11\times v^2=294.15[/tex]
v =7.31 m/sec
So velocity final velocity will be 7.31 m/sec