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At time t=0, a force F=k t where k is any constant, is exerted on a small body with mass m, which is initially at rest on a frictionless horizontal surface. The resultant direction of the applied force forms an angle of 90^{\circ} with the horizontal plane. Find the velocity of body at the moment of its breaking off plane.

Option: 1

0 \mathrm{~m} / \mathrm{sec}


Option: 2

30 \mathrm{~m} / \mathrm{sec}


Option: 3

20 \mathrm{~m} / \mathrm{sec}


Option: 4

25 \mathrm{~m} / \mathrm{sec}


Answers (1)


\mathrm{N}=\mathrm{mg}-\mathrm{ktsin} \alpha=0
\mathrm{t}=\frac{\boldsymbol{m g}}{\boldsymbol{k} \boldsymbol{\operatorname { s i n } \alpha}} \ldots(1)

From given force equation

\mathrm{m} \frac{d v}{d t}=\mathrm{kt} \cos \alpha \text {; }
Integrated

\begin{aligned} & \mathrm{m} \int_0^v \frac{d v}{d t}=\int_0^t kt \cos \alpha \\ & \mathrm{v}=\frac{k \cos \alpha}{2 m} t^2 \\ & \mathrm{~v}=0 \mathrm{~m} / \mathrm{sec}\left(\text { where }, \alpha=90^{\circ}\right) \end{aligned}

Posted by

Sumit Saini

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