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Two forces act on a 55-kg object. One force has magnitude 65 N directed 59° clockwise from the positive x-axis, and the other has a magnitude 35 N at 32° clockwise from the positive y-axis. What is the magnitude of this objectʚs acceleration? A) 1.1 m/s2 B) 1.3 m/s2 C) 1.5 m/s2

Respuesta :

Answer:

A) 1.1 m/s/s

Explanation:

There exist two forces on the object such that

[tex]F_1[/tex] = 65 N directed 59° clockwise from the positive x-axis

[tex]F_2[/tex] = 35 N at 32° clockwise from the positive y-axis

now we have

[tex]F_1 = 65 cos59\hat i - 65 sin59 \hat j[/tex]

[tex]F_2 = 35 sin32\hat i + 35 cos32 \hat j[/tex]

now the net force on the object is given as

[tex]F_{net} = F_1 + F_2[/tex]

[tex]F_{net} = (65 cos59 + 35 sin32)\hat i + (35cos32 - 65 sin59)\hat j[/tex]

[tex]F_{net} = 52\hat i - 26 \hat j[/tex]

so it's magnitude is given as

[tex]F_{net} = \sqrt{52^2 + 26^2} = 58.15 N[/tex]

now from Newton's II law we have

F = ma

[tex]a = \frac{58.15}{55} = 1.1 m/s^2[/tex]

onyebuchinnaji

The magnitude of this object's acceleration is 1.33 m/s².

The given parameters;

  • mass of the object, m = 55 kg
  • magnitude of first force, F₁ = 65 N, 59⁰
  • magnitude of the second force, F₂ = 35 N, 32⁰

The horizontal components of the two forces;

Fₓ = F₁cos(59) + F₂cos(32)

Fₓ = 65cos(59) + 35cos(32)

Fₓ = 33.48 + 29.68

Fₓ = 63.16

The vertical component of the two forces;

Fₙ = -F₁sin(59) + F₂sin(32)

Fₙ = -65sin(59) + 35sin(32)

Fₙ = -55.71 + 18.55

Fₙ = -37.16

The resultant of the two forces is calculated as follows;

[tex]\pi F = \sqrt{F_n^2 + F_x^2} \\\\F= \sqrt{(-37.16)^2 + (63.16)^2} \\\\F = \sqrt{5370.05} \\\\F = 73.28 \ N[/tex]

The magnitude of this object's acceleration is calculated using Newton's second law of motion;

F = ma

[tex]a = \frac{F}{m} \\\\a = \frac{73.28}{55} \\\\a = 1.33 \ m/s^2[/tex]

Thus, the magnitude of this object's acceleration is 1.33 m/s².

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