Question

An $8kg$ block of ice, released from rest at the top of a $1.5m$ long smooth ramp, slides down and falls with a velocity $2.5m{s}^{-1}$. Find the angle of the ramp with horizontal:
Note:

Sin (12) = 0.208

Sin (18) = 0.309

Sin (15) = 0.259

• A. ${12}^{\circ }$
• B. ${18}^{\circ }$
• C. ${15}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

Answer: (A)

${12}^{\circ }$

Let the ice block is coming down with acceleration $a$.
Use formula , $v^2-u^2=2aS$
here, $v=2.5m/s,u=0,S=1.5m$ so, ${2.5}^2=2a\left(1.5\right)\Rightarrow a=2.08m/s^2$
now, $ma=mg\sin \theta \Rightarrow \sin \theta =\frac{a}{g}=\frac{2.08}{10}=0.208$
$\therefore \theta ={\sin }^{-1}\left(0.208\right)={12}^0$