A box of mass $5\mathrm{kg}$ is pulled by a cord, up along a frictionless plane inclined at $30^\circ$ with the horizontal. The tension in the cord is $30\mathrm{N}$. The acceleration of the box is (Take$\mathrm{g}=10{\mathrm{ms}}^{-2}$ ):

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Correct answer: d) $1\mathrm{m}{\mathrm{s}}^{-2}$. $\begin{matrix}\text{Force acting up the incline (Tension, }T\text{):} & \ T & =30\text{ N} \ \text{Force acting down the incline:} & \ mg\sin ⁡\theta & =5\times 10\times \sin ⁡({30}^{∘}) \ & =50\times \frac{1}{2}=25\text{ N} \ \text{Net force equation (}{F}_{\text{net}}=ma\text{):} & \ T−mg\sin ⁡\theta & =ma \ 30−25 & =5\times a \ 5 & =5a \ a & =1\text{ m }{\text{s}}^{−2}\end{matrix}$

A box of mass \(5\mathrm{kg}\) is pulled by a cord, up along a frictionless plane inclined at \(30^\circ\) with the hori…

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