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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A 15 N force is applied to a $6.0\;kg$ block. Determine the acceleration of the block.

$\begin{array}{1 1} 1.146\;m/s^2 \\4.146 \;m/s^2\\8.14\;m/s^2 \\1.6\;m/s^2\end{array} $

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Answer : $1.146\;m/s^2$
Draw a free body diagram
At this point , We are as to the overall direction of motion , therefore we cannot correctly label $F_f$ on the diagram .
Step 2:
$F_{appx} =15 \cos 32=12.7207\;N$
$F_{appy} =15 \sin 32=7.9488\;N$
$F_{gx}=mg \sin \theta =(6.0)(9.80) \sin 32$
$\qquad= 31.1592 \;N$
$F_{appy}=15 \sin 32 =7.9488\;N$
$F_{gy} =mg \cos \theta =(6.0) (9.80) \cos 32$
$\qquad=49.8652\;N$
Step 3:
$F_N= F_{gy}+F_{appy }$
$\qquad= 49.8652+7.9488$
$\qquad= 57.8139\;N$
Therefore : $ F_f= \mu F _N$
$\qquad= (0.26) (57.8139\;N)$
$\qquad=11.5627 \;N$
Step 4:
We will call $F_f$ negative as the overall motion is down the ramp (in the direction of the force of gravity Since $F_{gx} > F_{app}$) , therefore the force of friction is directed in the opposite direction of the motion (up the ramp)
$F_{net}=F_{gx}-F_{appx} - F_{f}$
$\qquad=31.1592 \;N -12.7207 \;N -11.5627\;N =6.8758\;N$
$a=F_{net}/m =6.8758/6.0 =1.146\;m/s^2$
answered Aug 27, 2014 by meena.p
 

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