Respuesta :
Answer:
(a) Work done by the tension force is 987 J
(b) Coefficient of kinetic friction between the crate and surface is 0.495
Explanation:
(a)
Work done by any force F in moving an object by a distance d making an angle [tex]\Theta[/tex] with the direction of force is given by
[tex]W=Fd\cos (\Theta )[/tex]
[tex]W=175 \times 6 \times \cos (20 )J=987J[/tex]
Thus work done by the tension force is 987 J
(b)
Normal force on the crate is given by
[tex]N= mg-F\sin \Theta =(40\times 9.8-175\times \sin 20)Newton[/tex]
=>N=332 Newtons
Since crate is moving with constant speed . Therefore using Newtons second law .
[tex]Fcos(\Theta ) -\mu_k N=0[/tex]
Where [tex]\mu_k[/tex]=coefficient of kinetic friction
[tex]\therefore \mu_k=\frac{F\cos (\Theta )}{N}[/tex]
=>[tex]\mu_k=\frac{175\times \cos 20}{332}[/tex]
=>[tex]\mu _k= 0.495[/tex]
Thus coefficient of kinetic friction between the crate and surface is 0.495
