In Fig 8-33, a small block of mass $\mathit{m}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.032}\mathbf{\text{kg}}$can slide along the frictionless loop-the-loop, with loop radius $\mathit{R}\mathbf{=}\mathbf{}\mathbf{12}\mathbf{\text{cm}}$. The block is released from rest at point P, at height $\mathit{h}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.0}\mathit{R}$ above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point Pto a) point Q(b) the top of the loop?If the gravitational potential energy of the block–Earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is (c) at point P (d) at point Q (e) at the top of the loop? (f) If, instead of merely being released, the block is given some initial speed downward alongthe track, do the answers to (a) through (e) increase, decrease, or remain the same?

i) Mass of block $\text{m}=0.032\text{\hspace{0.17em}kg}$

ii) Loop radius $R=12\text{\hspace{0.17em}cm}=12\times {10}^{-2}\text{\hspace{0.17em}m}$

iii) Gravitational acceleration $g=9.8{\text{m/s}}^{\text{2}}$

iv) Height of block from the bottom $h=5R=60\times {10}^{-2}\text{m}$

The force of gravity is constant, so the work done can be found by using the formula for the work done in terms of gravitational force and displacement.

Formula:

Gravitational potential energy is given bytheformula,$U=mgh$

Gravitational force is mg$F=$mg

Work depends on displacement.

Displacement from point$P$to $Q$Pointis

$d=h-R$

$\Rightarrow d=5R-R$

$\Rightarrow d=4R$

$W_g=mgd$

$\Rightarrow W_g=mg4R$

$\Rightarrow W_g=0.032\times 9.8\times 4\times 12\times {10}^{-2}$

$\Rightarrow W_g=0.15\text{\hspace{0.17em}J}$

Work depends on displacement. Top loop is from 2R distance fromthebottom.

Displacement from pointto top of the loop is

$d=h-2R$

$\Rightarrow d=5R-2R$

$\Rightarrow d=3R$

$W_g=mgd$

$\Rightarrow W_g=mg4R$

$\Rightarrow W_g=0.032\times 9.8\times 3\times 12\times {10}^{-2}$

$\Rightarrow W_g=0.11\text{\hspace{0.17em}J}$

The potential energy at pointis

$U=mgh$

$\Rightarrow U=5mgR$

$\Rightarrow U=0.032\times 9.80\times 5\times 12\times {10}^{-2}$

$\Rightarrow U=0.19\text{\hspace{0.17em}J}$

At point Q total height is$R$

The potential energy at point$Q$is

$U=mgR$

$\Rightarrow U=0.032\times 9.80\times 12\times {10}^{-2}$

$\Rightarrow U=0.038\text{\hspace{0.17em}J}$

At top of the loop total height is

The potential energy at top of the loop is

$U=2mgR$

$\Rightarrow U=0.032\times 9.80\times 12\times {10}^{-2}$

$\Rightarrow U=0.075\text{\hspace{0.17em}J}$

The potential energy or work does not depend on the initial speed so, if initial speed $V_i\ne 0$then the result a) and e) is unchanged.