Equations of Motion: Practice Problem & Solution
A body of mass \( m \) is projected along a rough inclined plane (having an angle of inclination with horizontal \( \theta \), equal to angle of repose) with a velocity \( v \). It travels up a maximum distance \( s \) before it comes to a halt. Then \( v \) is:
Solution Explained:
To solve this problem, we apply the core principles of Equations of Motion. Understanding the underlying formula is key to arriving at the correct answer below:
Since the inclination equals the angle of repose, \( \mu = \tan\theta \). The acceleration down the incline during upward motion is \( a = g \sin\theta + \mu g \cos\theta = 2g \sin\theta \). Using \( v^2 = 2as \), we get \( v = 2\sqrt{gs \sin\theta} \).
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