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Vika [28.1K]
3 years ago
5

Part 1 :

Physics
1 answer:
dybincka [34]3 years ago
7 0

Answer:

1. 218.55 N

2. 30.96^{o}

3. 2.1 m/s^{2}

Explanation:

Part 1;

Net force F=mg sin \theta where m is mass, g is gravitational force and \theta is the angle of inclination

F= 46*9.8*sin 29^{o}= 218.55N

Frictional force, F_{r} is given by

F_{r} = \mu_{s}mg cos \theta where \mu_{s} is the coefficient of static friction

F_{r} = 0.6*46*9.8 cos 29

F_{r}=236.57N

Since F_{r}>F, therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force F_{s} = \mu_{s} mg cos \theta

mg sin \theta =\mu_{s} mg cos \theta

\mu_{s}= \frac {sin \theta}{cos \theta}

\mu_{s}= tan \theta

The maximum angle of inclination \theta = tan^{-1} \mu_{s}

\theta = tan^{-1} (0.6)

\theta= 30.96^{o}  

        

Part 3:

Net force on the object is given by

ma = mg sin 38 - \mu_{k} mg cos 38 where \mu_{k} is the coefficient of kinetic friction

 a = g ( sin 38 - \mu_{k} cos 38 )

                 = 9.8 ( sin 38 - (0.51) cos 38 )

                = 2.1m/s^{2}

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