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PART - 1 CH - 4 Laws of Motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 1 CH - 4 Laws of Motion2 Marks
Question
A 2 kg block is kept at rest on a flat surface. A horizontal force $F$ on the block is gradually increased. When the value of $F$ becomes 8 N the block starts moving. Once the motion starts, it starts moving at a uniform speed only with the help of 7N force. Find out :
(i) Static and dynamic friction coefficient.
(ii) When the value of $F$ is 5 N , then the static friction force acting on the block is.
(iii) When the value of $F$ is again 8 N the acceleration of moving block. ( $g = 9 . 8 ~ m / s ^2$ )
✓

Answer

 It is given :
$\text { (i) } \begin{aligned}
\text { Mass of block }(m) & =2 kg \\
\text { Static friction force } F_s & =m g \mu_s \\
8 & =2 \times 10 \mu_s \\
\therefore \quad & \mu_s \\
\therefore \quad & =\frac{8}{20}=\frac{4}{10}=0.4 \\
\mu_s & =0.4\quad \text { Ans. }
\end{aligned}$
Dynamic/kinetic friction force $\left( F _k\right)$
$\begin{aligned}
& =\mu_k m g \\
7 & =2 \times 10 \mu_k \\
\mu_k & =\frac{7}{20}=0.35\quad \text { Ans. }
\end{aligned}$
(ii) The magnitude of static friction force is equal to applied force but the directions are opposite. So on applying force of 5 N ,
$\therefore \quad$ Static friction force $=5 N\quad \text { Ans. }$
(iii) When a force of 8 N is applied on a moving block the force on the block $=($ applied force $)-($ friction force on moving object)
$=8 N-7 N=1 N$
$\therefore$ Acceleration of moving block
$\begin{aligned}
a & =\frac{F}{m} \\
& =\frac{1}{2}=0.5 m / s^2\quad \text { Ans. }
\end{aligned}$ 

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