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]
The magnitude of this object's acceleration is 1.33 m/s².
The given parameters;
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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