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4-2.Friction
hard
Find the acceleration of $a_1, a_2$ and $a_3$. If $10\,N$ force $F$ is applied on $3\,kg$ block :-
A$a_1 = 3\,m/s^2, a_2 = a_3 = .4 \,m/s^2$
B$a_1 = a_2 = a_3 = \frac{5}{6}\, m/s^2$
C$a_1 = 3\,m/s^2, a_2 = a_3 = 1 \,m/s^2$
DNone
Solution
$F=10 \mathrm{N}$
$\text { Friction on } 3 \mathrm{kg} \quad \mathrm{F}_{2}=0.3 \times 5 \times 10=15 \mathrm{N}$
$\begin{aligned} \text { acco. of system } a &=\frac{F}{\mathrm{m}} \\ &=\frac{10}{12}=\frac{5}{6} \end{aligned}$
Pseudo on $2 \mathrm{kg}$ $\mathrm{F}^{\prime}=2 \times \frac{5}{6}=\frac{5}{3} \mathrm{N}$
$\text { Friction on } 2 \mathrm{kg} \quad \mathrm{F}_{1}=0.2 \times 2 \times 10=4 \mathrm{N}$
So all blocks move together by $\frac{5}{6} \mathrm{m} / \mathrm{s}^{2}$
$\text { Friction on } 3 \mathrm{kg} \quad \mathrm{F}_{2}=0.3 \times 5 \times 10=15 \mathrm{N}$
$\begin{aligned} \text { acco. of system } a &=\frac{F}{\mathrm{m}} \\ &=\frac{10}{12}=\frac{5}{6} \end{aligned}$
Pseudo on $2 \mathrm{kg}$ $\mathrm{F}^{\prime}=2 \times \frac{5}{6}=\frac{5}{3} \mathrm{N}$
$\text { Friction on } 2 \mathrm{kg} \quad \mathrm{F}_{1}=0.2 \times 2 \times 10=4 \mathrm{N}$
So all blocks move together by $\frac{5}{6} \mathrm{m} / \mathrm{s}^{2}$
Standard 11
Physics
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