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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A small car with a mass of $0.8\; kg$ travels at constant speed on the inside of a track that is a vertical circle with a radius of $5.0\; m.$ If the normal force exerted by the track on the car when it is at the top of the track is $6.0\; N$, what is the normal force on the car when it is at the bottom of the track?

$\begin{array}{1 1} 21.7\;N \\ 25.8\;N \\ 2.7\;N \\3.6\;N \end{array} $

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Answer : 21.7 N
Step 1:
Two forces are acting on the car , gravity and the normal force.
At the top, both forces are toward the center of the circle, so Newton’s second law gives $m_g= n_B= ma_{rad}$
At the bottom , gravity is downward but the normal force is upward. so $n_A-mg= ma_{rad}$
Solving for the acceleration
$a_{rad} =g+ \large\frac{n_B}{m}$
$\qquad= 9.8 m/s +\large\frac{6N}{0.8\;kg}$
$\qquad= 17.3 m/s^2$
Step 2:
The normal force can now be calculated:
$n_A= m(g+a_{rad})$
$\qquad= 0.8 \;kg (9.8 +17.3)m/s^2$
$\qquad= 21.7 \;N$
answered Aug 25, 2014 by meena.p

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