Laws Of Motion Practice Questions

Assuming acceleration $a$ of the man is downwards. So the equation will be $m g-T=m a$ $\therefore a=g-\;\frac{T}{m}=10-\;\frac{360}{60}=4 \frac{\text{m}}{\text{s}}^2$ So the maximum acceleration of man is $4 \frac{\text{m}}{\text{s}}^2$ downwards.

Let the two forces be $F_1$ and $F_2$ and let $F_2$ is smaller than $F_1$ and assume $R$ is the resultant force. Given $F_1+F_2=18$ .....(i) From the right angle triangle, $F_2^{2+}R^2=F_1^2$ or $F_1^{2-}F_2^2=R^2$ or $(F_1+F_2)(F_1-F_2)=R^2$ or $(18)(F_1-F_2)=(12)^2=144$ or $(F_1-F_2)=8$.....(ii) By solving equation $(i)$ and $(i i)$ we get, $F_1=13 N$ and $F_2=5 {\text{N}}$

Assume tension between block $A$ and $B$ is $T_1$ and block $B$ and $C$ is $T_2$. Acceleration $a=0.6 {\text{m}} / {\text{s}}^2$. For block A : ${\text{F}}- {\text{T}}_1= {\text{ma}}$ For block B : ${\text{T}}_1- {\text{T}}_2= {\text{ma}}$ By adding both those equation we get, ${\text{F}}- {\text{T}}_2=2 {\text{ma}}$ ${\text{T}}_2= {\text{F}}-2 {\text{ma}}$ $T_2=10.2-2 \times 2 \times 0.6=10.2-2.4=7.8 {\text{N}}$

Let acceleration of ball $=\vec{a}_b$ and acceleration of man is $=\vec{a}_m$ With respect to the man standing in the lift, the acceleration of the ball $\vec{a}_{b m}=\vec{a}_b-\vec{a}_m$ $\Rightarrow a_{b m}=g-a$ Where $a$ is the acceleration of the man as the acceleration of the lift is $a$. With respect to the man standing on the ground the acceleration of the ball $\vec{a}_{b m}=\vec{a}_b-\vec{a}_m$ $\Rightarrow a_{b m}=g-0$ $=g$

When $F_1, F_2$ and $F_3$ are acting on a particle then the particle remains stationary. This means that the resultant of $F_1, F_2$ and $F_3$ is zero. When $F_1$ is removed then particle will start moving due to the force $F_2$ and $F_3$ in the resultant of $F_2$ and $F_3$ and it should be equal and opposite to $F_1$. i.e. $|\vec{F_2}+\vec{F_3}|=|\vec{F_1}|$ $\therefore \;\; a=\;\frac{|\vec{F_2}+\vec{F_3}|}{m} \Rightarrow a=\;\frac{F_1}{m}$