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QJEEM23Phec07: JEE Main 2023 Physics - Class 11 JEEM23Phec07 Solution: An object of mass 'm' initially at rest on a smooth horizontal…

An object of mass 'm' initially at rest on a smooth horizontal plane starts moving under the action of force F = 2N. In the process of its linear motion, the angle θ\theta (as shown in figure) between the direction of force and horizontal varies as θ=kx\theta=\mathrm{k}x , where k is a constant and xx is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be E=nksin⁡θE = {n \over k}\sin \theta . The value of n is ___________.

JEE Main 2023 (Online) 25th January Morning Shift Physics - Work Power & Energy Question 45 English

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JEE Main 2023 (Online) 25th January Morning Shift Physics - Work Power & Energy Question 45 English Explanation

 Work done =ΔK.E\text { Work done }=\Delta \mathrm{K} . \mathrm{E}

amp;∴∫F⋅dx=12mv2=Eamp;∴E=∫0x2cos⁡(kx)dxamp;E=2k[sin⁡kx]0xamp;=2ksin⁡kxamp;=2sin⁡θk\begin{aligned}& \therefore \int F \cdot d x=\frac{1}{2} m v^{2}=E \\\\& \therefore E=\int_{0}^{x} 2 \cos (k x) d x \\\\& E=\frac{2}{k}[\sin k x]_{0}^{x} \\\\& =\frac{2}{k} \sin k x \\\\& =\frac{2 \sin \theta}{k}\end{aligned}

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