An object with mass \(m\) kilograms slides down a smooth inclined plane with velocity \( \underset{\sim}{v}(t)\), where \(t\) is the time in seconds after the object started sliding down the plane. The inclined plane makes an angle \(\theta\) with the horizontal, as shown in the diagram. The normal reaction force is \(\underset{\sim}{R}\). The acceleration due to gravity is \(\underset{\sim}{g}\) and has magnitude \(g\). No other forces act on the object.
The vectors \(\underset{\sim}{i}\) and \( \underset{\sim}{j} \) are unit vectors parallel and perpendicular, respectively, to the plane, as shown in the diagram.
- Show that the resultant force on the object is \(\underset{\sim}{F}=-(m g \ \sin \theta) \underset{\sim}{i}\). (2 marks)
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- Given that the object is initially at rest, find its velocity \(\underset{\sim}{v}(t)\) in terms of \(g\), \(\theta, t\) and \(\underset{\sim}{i}\). (2 marks)
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i.
\(\text{Resolving forces in}\ \underset{\sim}{j} \ \text{direction:} \)
\( {\underset{\sim}{F}}_\underset{\sim}{j} = \underset{\sim}{R} + m\underset{\sim}{g}\ \cos \theta = 0\ \ \text{(in equilibrium)} \)
\(\text{Resolving forces in}\ \underset{\sim}{i} \ \text{direction:} \)
\( {\underset{\sim}{F}}_\underset{\sim}{i} = -m\underset{\sim}{g} \ \sin \theta \ \ \ \text{(down slope)} \)
\(\therefore \text{Resultant force:}\ \ \underset{\sim}{F}=-(m g \ \sin \theta) \underset{\sim}{i} \)
ii. \(\text{Using}\ \ \underset{\sim}{F}=m \underset{\sim}{a}: \)
| \(m \underset{\sim}{a}\) | \(=-mg\ \sin \theta \ \underset{\sim}{i} \) | |
| \(\underset{\sim}{a}\) | \(=-g\ \sin \theta \ \underset{\sim}{i} \) | |
| \(\underset{\sim}{v}\) | \(= \displaystyle \int \underset{\sim}{a}\ dt \) | |
| \(=-gt\ \sin \theta +c \) |
\(\text{When}\ \ t=0,\ \ \underset{\sim}{v}=0\ \ \Rightarrow \ \ c=0 \)
\(\therefore \underset{\sim}{v}=-gt\ \sin \theta \ \underset{\sim}{i} \)