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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

Determine the rate of deceleration of $7.0 \;kg $ block sliding up an incline angled at $35^{\circ}$ to the horizontal. Assume that the coefficient of friction is $0.45$

$\begin{array}{1 1}-9.2\;m/s^2\\6.7\;m/s^2\\7.5 \;m/s^2\\10\;m/s^2\end{array} $

1 Answer

Step 1:
Free Body Diagram
Step 2:
The component of gravity , $F_g$
$F_{gx} =F_g \sin \theta$
$F_{gx}=mg \sin \theta$
$\qquad= (7.0)(9.80) \sin 35$
$\qquad= 39.3473\;N$
Step 3:
The force of friction , $F_f$
$F_f= \mu _k F_N$
$F_f = \mu _k mg \cos \theta$
$\qquad= (0.45) (7.0)(9.80) \cos 35$
$\qquad= 25.2872\;N$
Step 4:
In this case, both the forces are opposite to the direction of motion, and are therefore both negative.
$F_{net} =-F_{gx} -F_f$
$F_{net }= -39.3473 \;N- 25.2872=-64.6345$
$a= F_{net/m}=-64.6345/7.0$
$a= F_{net /m} =-64.6345/7.0 =-9.2 m/s^2$
answered Aug 27, 2014 by meena.p
 

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