Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
0 votes

An $8.0\;kg$ box is released on a $30^{\circ}$ inclined and accelerates down the incline at $30\;m/s^2$ Find the frictional force impending its motion . How large is the coefficient of friction in this situation ?

$\begin{array}{1 1}2.76\\0.54\\24 \\62\end{array} $

Can you answer this question?

1 Answer

0 votes
Answer : 0.54
The component of the weight down the incline $=8(9.8) (0.5) =39.2 \;N$
Now $F=ma$ leads to $39.2 -f=8(0.3)$, So that $f= 36.8 \;N$ .
The Normal force equals the component of W perpendicular to the incline , $8(9.8)(0.867)=67.9\;N$.
Therefore $\mu =36.8/67.9 =0.54$
answered Aug 28, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App