A body of mass 1 kg is acted upon by two perpendicular forces of 3 N and 4 N. What is the magnitude of the acceleration of the body?

A. 5 m/s^2 — The resultant force is the vector sum of the two perpendicular forces, F_net = sqrt(3^2 + 4^2) = 5 N. Since F = ma, the acceleration is a = 5 / 1 = 5 m/s^2.

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$A body of mass m = 1 kg is acted upon by two perpendicular forces of F_{1} = 3 N and F_{2} = 4 N . What is the magnitude of the acceleration a of the body?$

A. $5 m/s^{2}$
B. $7 m/s^{2}$
C. $1 m/s^{2}$
D. $3.5 m/s^{2}$

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Correct answer: A — $5 m/s^{2}$

Solution

The resultant force is F_{net} = F_{1}^{2} + F_{2}^{2} = 3^{2} + 4^{2} = 5 N . Since F = ma, the acceleration is a = \frac{F _{net}}{m} = \frac{5}{1} = 5 m/s^{2} .

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A body of mass m=1 kg is acted upon by two perpendicular forces of F1​=3 N and F2​=4 N. What is the magnitude of the acceleration a of the body?

Solution

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More Laws of Motion practice questions

$A body of mass m = 1 kg is acted upon by two perpendicular forces of F_{1} = 3 N and F_{2} = 4 N . What is the magnitude of the acceleration a of the body?$