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A box weighing 46 newtons rests on an incline that makes an angle of 25° with the horizontal. What is the magnitude of the component of the box’s weight perpendicular to the incline?
(1)19 N
(2)21 N
(3)42 N
(4)46 N

Respuesta :

MissPhiladelphia

The magnitude of the component of the box’s weight perpendicular to the incline can be olve using the formula:

F = wcos(a)

Where F is the box’s weight perpendicular to the incline

W is the weight of the box

A is the angle of the incline

F = (46)cos(25)

F = 42 N

snehashish65

Answer:

The magnitude of the component of the box’s weight perpendicular to the incline is 42 N. Hence, option (3) is correct.

Explanation:

Given data:

Weight of box is, [tex]W = 46 \;\rm N[/tex].

Angle of inclination is, [tex]\theta = 25^{\circ}[/tex].

In an inclined plane, the weight of box perpendicular to the incline is the cosine component of weight. Then,

[tex]W'=Wcos \theta[/tex]

Substituting the values as,

[tex]W'=46 \times cos 25^{\circ}\\W' \approx 42 \;\rm N[/tex]

Thus, the magnitude of the component of the box’s weight perpendicular to the incline is 42 N.

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