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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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Consider a block of mass 5 kg on a table top initially at rest. Assume that the surface is frictionless. Let a pushing force of 50 N directed at an angle of 30$^\circ$ below the horizontal act on the block. Find the acceleration of the block and the normal force that the table exerts on the block.


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Answer: 8.66 ms$^{-2}$ and 74 N
Here is the force diagram for the problem.
According to Newton's 2nd law, $F = ma_x = F_P \cos 30^\circ$
Given that $F_P = 50N,\;$ and $m = 5\; Kg \rightarrow a_x = \large\frac{50N \times \cos 30 ^\circ}{5 kg}$$ = 8.66 m/s^2$
From the diagram above, we see that $N - mg - F_P \sin30^\circ = 0$
$\Rightarrow N = mg + F_P \sin 30^\circ = 5 \times 9.8 + 50 \sin 30 = 74 \;N$
answered Aug 19, 2014 by balaji.thirumalai
edited Aug 19, 2014 by balaji.thirumalai
 

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