Question

A block slides along a track from one level to another level, by moving through an intermediate valley. Find the maximum value of $h$ (height) that the block can reach, if the initial speed of the block is $v_0$. (Assume track is frictionless)

• A. $2g(d+\frac{v_0^2}{2g})$
• B. $\frac{1}{2g}(2gd+v_0^2)$
• C. $\frac{v_o^2}{2g}$
• D. $2gd$

Answer 1

The correct option is B $\frac{1}{2g}(2gd+v_0^2)$

Initial speed of block $\left(u\right)=v_0$
Final speed of block $=0$
Speed of block when it reaches point $B$
From conservation of energy principle,
${\left(TE\right)}_A={\left(TE\right)}_C$
$mgd+\frac{1}{2}m{v}_0^2=mgh+\frac{1}{2}m(0{)}^2$
$\Rightarrow mgh=\frac{1}{2}m(2gd+v_0^2)$
$\Rightarrow h=\frac{1}{2g}(2gd+v_0^2)$