Physics
10.01.2023 09:29
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If the only forces acting on a 2.0kg mass are f1 = (3i-8j)n and f2 = (5i+3j)n, what

If the only forces acting on a 2.0kg mass are f1 = (3i-8j)n and f2 = (5i+3j)n, what is the magnitude of the acceleration of the particle?
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kobiemajak
kobiemajak
4,8(72 marks)

The magnitude of the object's acceleration is 4.72 m/s².

The given parameters;

mass of the object, m = 2 kgfirst force on the object, F1 = (3i - 8j) Nsecond force on the object, F2 = (5i +3j)

The net force on the object is calculated as follows;

F_{net} = F_1 + F_2\\\\F_{net} = (3i \ - \ 8j)\ + \ (5i \ + \ 3j)\\\\F_{net} = 8i \ - 5j

The resultant force on the object is calculated as follows;

|F| = \sqrt{(8)^2 + (-5)^2}  \\\\|F| = 9.43 \ N\\\\

The magnitude of the object's acceleration is calculated by applying Newton's second law of motion;

F = ma\\\\a = \frac{F}{m} \\\\a = \frac{9.43}{2} \\\\a = 4.72\ m/s^2

Thus, the magnitude of the object's acceleration is 4.72 m/s².

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chevi2005
chevi2005
5,0(29 marks)

4.7m/s²

Explanation:

According to newton's first law,

Force = mass × acceleration

Since we are given more the one force, we will take the resultant of the two vectors.

Mass = 2.0kg

F1+F2 = (3i-8j)+(5i+3j)

Adding component wise, we have;

F1+F2 = 3i+5i-8j+3j

F1+F2 = 8i-5j

Resultant of the sum of the forces will be;

R² = (8i)²+(-5j)²

Since i.i = j.j = 1

R² = 8²+5²

R² = 64+25

R² = 89

R = √89

R = 9.4N

Our resultant force = 9.4N

Substituting in the formula

F = ma

9.4 = 2a

a = 9.4/2

a = 4.7m/s²

Therefore, magnitude of the acceleration of the particle is 4.7m/s²

charlesiarenee0
charlesiarenee0
4,8(76 marks)

yoy

Explanation:

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