Question 12 Marks
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg , (Neglect air resistance throughout.) Just after it is dropped from the window of a train accelerating with $1 \mathrm{~ms}^{-2}$.
Answer
View full question & answer→1 N ; vertically downward It is given that the train is accelerating at the rate of $1 \mathrm{~m} / \mathrm{s}^2$. Therefore, the net force acting on the stone, $\mathrm{F}^{\prime}=\mathrm{ma}=0.1 \times 1=0.1 \mathrm{~N}$ This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force F', stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations. Therefore, the net force acting on the stone is given only by acceleration due to gravity. $\mathrm{F}=\mathrm{mg}=1 \mathrm{~N}$ This force acts vertically downward.
Mg - T = Ma T - mg = ma Solving the equation, we get $\text{a}=\frac{(\text{M}-\text{m})\text{g}}{\text{M}+\text{m}}$ and $\text{T}=\frac{(2\text{Mm})\text{g}}{\text{M}+\text{m}}$
$\text{F}=\text{N} $ and $\text{f}=\text{Mg}$$\Rightarrow\mu\text{F}=\text{Mg}$ or $\text{F}=\frac{\text{Mg}}{\mu}$ is the minimum force to hold the block against the wall at rest.
i.e., $\text{Mg}\sin\theta=\text{Ma}\Rightarrow\text{a}=\text{g}\sin\theta$ Also, $\text{Mg}\cos\theta-\text{N}=0$ $\Rightarrow\text{N}=\text{Mg}\cos\theta.$
$\therefore\ \text{N}\sin\theta=\frac{\text{mv}^2}{\text{r}},$ $\text{N}\cos\theta=\text{mg}$ Dividing, $\tan\theta=\frac{\text{v}^2}{\text{rg}}$ $\Rightarrow\ \theta=\tan^{-1}\Big(\frac{\text{v}^2}{\text{rg}}\Big)$