Question

The three forces (in units of N) given below are acting on a 20 kg mass. Calculate the magnitude of the acceleration of the mass. stack F subscript 1 with rightwards harpoon with barb upwards on top equals 3 i with hat on top space space space space space stack F subscript 2 with rightwards harpoon with barb upwards on top equals 5 j with hat on top space space space space stack F subscript 3 with rightwards harpoon with barb upwards on top equals open parentheses i with hat on top minus 3 j with hat on top close parentheses space space A. 0.2 m/s^2 B. 0.224 m/s^2 C. 0.1 m/s^2 D. 1.0 m/s^2

Answer 1

Answer:

B. 0.224 m/s²

Explanation:

Given:

Mass of the object (m) = 20 kg

The forces acting on the object are:

$\vec{F_1}=3\vec{i}\ N\\\\\vec{F_2}=5\vec{j}\ N\\\\\vec{F_3}=(\vec{i}-3\vec{j})\ N$

Now, the net force acting on the object is equal to the vector sum of the forces acting on it. Therefore,

$\vec{F_{net}}=\vec{F_1}+\vec{F_2}+\vec{F_3}\\\\\vec{F_{net}}=3\vec{i}+5\vec{j}+\vec{i}-3\vec{j}\\\\\vec{F_{net}}=(3+1)\vec{i}+(5-3)\vec{j}\\\\\vec{F_{net}}=(4\vec{i}+2\vec{j})\ N$

Now, the magnitude of the net force is equal to the square root of the sum of the squares of its components and is given as:

$|\vec{F_{net}}|=\sqrt{4^2+2^2}\\\\|\vec{F_{net}}|=\sqrt{20}\ N$

Now, from Newton's second law, the magnitude of acceleration is equal to the ratio of the magnitude of net force and mass. So,

Magnitude of acceleration is given as:

$|\vec{a}|=\dfrac{|\vec{F_{net}}|}{m}\\\\|\vec{a}|=\frac{\sqrt{20}\ N}{20\ kg}\\\\|\vec{a}|=0.224\ m/s^2$

Therefore, option (B) is correct.