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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

Verified Answer :

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, \(W = 46 \;\rm N\).

Angle of inclination is, \(\theta = 25^{\circ}\).

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

\(W'=Wcos \theta\)

Substituting the values as,

\(W'=46 \times cos 25^{\circ}\\W' \approx 42 \;\rm N\)

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

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