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JEE Mains - Work, Power & Energy
Question: An object of mass \( m \) initially at rest on a smooth horizontal plane starts moving under the action of a constant force \[ F = 2\,\text{N}. \] In the process of its linear motion, the angle \( \theta \) (as shown in the figure) between the direction of the force and the horizontal varies as \[ \theta = kx, \] where \( k \) is a constant and \( x \) is the distance covered by the object from its initial position. The kinetic energy of the object is expressed as \[ E = \frac{n}{k}\sin \theta. \] Determine the value of \( n \).
[JEE Mains Physics]
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Solution
Since the object moves on a smooth horizontal plane, the work done by the force goes entirely into its kinetic energy. The force \( F = 2\,\text{N} \) acts at an angle \( \theta \) with the horizontal, so only its horizontal component \( F\cos\theta \) does work.
An infinitesimal work done over a displacement \( dx \) is: \[ dW = F\cos\theta\,dx = 2\cos\theta\,dx. \] Given that the angle varies as: \[ \theta = kx, \] we substitute to obtain: \[ dW = 2\cos(kx)\,dx. \]
The total work done (equal to the kinetic energy \( E \)) from \( x = 0 \) to \( x \) is: \[ E = \int_{0}^{x} 2\cos(kx)\,dx. \] To integrate, substitute \( \theta = kx \) so that \( d\theta = k\,dx \) or \( dx = \frac{d\theta}{k} \). The limits then change from \( \theta = 0 \) to \( \theta \): \[ E = \frac{2}{k}\int_{0}^{\theta} \cos\theta\,d\theta = \frac{2}{k}\sin\theta. \]
Comparing this with the given expression \( E = \frac{n}{k}\sin \theta \), we deduce that: \[ n = 2. \]
