JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
An object of mass ' \(m\) ' initially at rest on a smooth horizontal plane starts moving under the action of force \(F=2 N\). In the process of its linear motion, the angle \(\theta\) (as shown in figure) between the direction of force and horizontal varies as \(\theta= kx\), where \(k\) is a constant and \(x\) is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be \(E =\frac{ n }{ k } \sin \theta\). The value of \(n\) is \(.....\)

- A \(1\)
- B \(3\)
- C \(4\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(F \cos \theta= ma\) \(2 \cos ( kx )=\frac{ mvdv }{ dx }\) \(\int \limits_0^v v d v=2 \int \limits_0^x \cos (k x) d x\) \(K . E .=\frac{2}{ k } \sin \theta\) \(n=2\)
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