A force of 150 N is continuously applied to an object with a mass of 11 kg over a distance of 2.5 m.The object slides over a surface with a coefficient of kinetic friction of .3.How fast is the object moving immediately after the force ceases to be applied?

Respuesta :

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

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