A box of mass 5 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 N. The acceleration of the box is (Take $g=10\ m\,s^{-2}$)

A 2 m s$^{-2}$

B Zero

C 0.1 m s$^{-2}$

D 1 m s$^{-2}$

The answer is D.

Solution

$F_{net}=Ma=T-Mg\sin30^\circ$. $30-5\times10\times\dfrac{1}{2}=5a\Rightarrow 30-25=5a\Rightarrow a=1\ m\,s^{-2}$.

Concept Motion on a frictionless inclined plane

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