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

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 speed of the car?

$\begin{array}{1 1} 21.7\;m/s \\ 9.3\;m/s \\ 2.7\;m/s \\3.6\;m/s \end{array} $

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Answer : 9.3 m/s
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 $mg+n_B=ma_{rad}$
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 speed of the car can now be calculated :
$v= \sqrt {a_{rad} R}$
$\quad= \sqrt {17.3 m/s^2.5 \;m}$
$\qquad= 9.3 m/s$
answered Aug 25, 2014 by meena.p

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