Question 75: Laws of Motion

Question

Three forces $$F_{1}=10 \mathrm{~N}, F_{2}=8 \mathrm{~N}, \mathrm{~F}_{3}=6 \mathrm{~N}$$ are acting on a particle of mass $$5 \mathrm{~kg}$$. The forces $$\mathrm{F}_{2}$$ and $$\mathrm{F}_{3}$$ are applied perpendicularly so that particle remains at rest. If the force $$F_{1}$$ is removed, then the acceleration of the particle is:

Accepted Answer

Since the particle is initially at rest, the net force acting on it is zero. This means that the forces $$F_1$$, $$F_2$$, and $$F_3$$ are balanced. Given that $$F_2$$ and $$F_3$$ are acting perpendicularly, we can represent the balance of forces as follows:
$$F_1 = \sqrt{F_2^2 + F_3^2}$$
We can now calculate the equivalent force resulting from $$F_2$$ and $$F_3$$:
$$F_1 = \sqrt{(8 \mathrm{~N})^2 + (6 \mathrm{~N})^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \mathrm{~N}$$
Since the particle is initially at rest, when the force $$F_1$$ is removed, only $$F_2$$ and $$F_3$$ remain. The net force acting on the particle is the equivalent force resulting from $$F_2$$ and $$F_3$$, which we have calculated to be $$10 \mathrm{~N}$$.
Now, we can use Newton's second law to find the acceleration of the particle:
$$F = ma$$
Where:
$$F$$ is the net force acting on the particle
$$m$$ is the mass of the particle
$$a$$ is the acceleration of the particle
Rearranging the equation to solve for $$a$$:
$$a = \frac{F}{m}$$
Plugging in the values for $$F$$ and $$m$$:
$$a = \frac{10 \mathrm{~N}}{5 \mathrm{~kg}} = 2 \mathrm{~ms}^{-2}$$
The acceleration of the particle when the force $$F_1$$ is removed is $$2 \mathrm{~ms}^{-2}$$, which corresponds to Option C.

Answer

$$4.8 \mathrm{~ms}^{-2}$$

Answer

$$7 \mathrm{~ms}^{-2}$$

Answer

$$2 \mathrm{~ms}^{-2}$$

Answer

$$0.5 \mathrm{~ms}^{-2}$$

Laws of Motion

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