A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If $$\vec{F_1}$$ is the force required to just move the block up the inclined plane and $$\vec{F_2}$$ is the force required to just prevent the block from sliding down, then the value of $$|\vec{F_1}| - |\vec{F_2}|$$ is: [Use $$g = 10$$ m s$$^{-2}$$]

$$m=5kg,θ=30°,μ=0.1=1/10,g=10$$

Normal reaction:
$$N=mg\cosθ=5\times10\times\cos30°=50\times(\sqrt{3}/2)=25\sqrt{3}$$

Friction:
$$f=μN=(1/10)\times25\sqrt{3}=(5\sqrt{3})/2$$

Now case 1: just moving up the plane

Friction acts down the plane

$$F₁=mg\sinθ+f$$
$$=5\times10\times(1/2)+(5\sqrt{3})/2$$
$$=25+(5\sqrt{3})/2$$

Now case 2: just preventing sliding down

Friction acts up the plane

$$F₂=mg\sinθ−f$$
$$=25−(5\sqrt{3})/2$$

Now difference:

$$|F₁|−|F₂|$$
$$=[25+(5\sqrt{3})/2]−[25−(5\sqrt{3})/2]$$

$$\frac{5\sqrt{3}}{2}-\left(-\frac{5\sqrt{3}}{2}\right)=\frac{5\sqrt{3}}{2}+\frac{5\sqrt{3}}{2}$$

$$=5\sqrt{3}$$

Final answer:

$$=5\sqrt{3}$$