Question

an object 16n of force being applied to the right 126n of force being applied to the left and 4n force being applied downward what is the acceleration of an object if its mass was 0.35kg

Answer 1

The acceleration of the object is 314.57 m/
`s^(2)`.

Step-by-step explanation:

From Newton's second law of motion,

F = ma

where: F is the force applied on an object, m is the mass of the object and a is the acceleration of the object.

⇒ a =
`(F)/(m)`

In the given question, the resultant force, F should be determined.

Total horizontal force,
`F_(x)` = 16 - 126

= -110 N

The horizontal force applied to the object is 110 N to the left.

Vertical force,
`F_(y)` = 4 N.

Resultant force, F =
`\sqrt{F_(x) ^(2) + F_(y) ^(2) }`

=
`\sqrt{(-110)^(2) + (4)^(2) }`

=
`√(12116)`

= 110.0727

Resultant force, F = 110.1 N

So that,

a =
`(110.1)/(0.35)`

= 314.5714

a = 314.57 m/
`s^(2)`

The acceleration of the object is 314.57 m/
`s^(2)`.