Laws Of Motion Question 10
Question 10 - 31 January - Shift 1
As shown in figure, a $70 kg$ garden roller is pushed with a force of $\overrightarrow{{}F}=200 N$ at an angle of $30^{\circ}$ with horizontal. The normal reaction on the roller is (Given $g=10 m s^{-2}$ )
(1) $800 \sqrt{2} N$
(2) $600 N$
(3) $800 N$
(4) $200 \sqrt{3} N$
Show Answer
Answer: (3)
Solution:
Formula: Second Law Of Motion
$N=mg+F \sin 30^{\circ}$
$=700+200 \times \frac{1}{2}=800$ newton.
