91Ó°ÊÓ

i) Mass of block $\text{m}=0.032\text{ k²µ}$

ii) Loop radius $R=12\text{ c³¾}=12×{10}^{−2}\text{″¾}$

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

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

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

$d=h−R$

$⇒d=5R−R$

$⇒d=4R$

$W_g=mgd$

$⇒W_g=mg4R$

$⇒W_g=0.032×9.8×4×12×{10}^{−2}$

$⇒W_g=0.15\text{ J}$

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

Displacement from pointto top of the loop is

$d=h−2R$

$⇒d=5R−2R$

$⇒d=3R$

$W_g=mgd$

$⇒W_g=mg4R$

$⇒W_g=0.032×9.8×3×12×{10}^{−2}$

$⇒W_g=0.11\text{ J}$

The potential energy at pointis

$U=mgh$

$⇒U=5mgR$

$⇒U=0.032×9.80×5×12×{10}^{−2}$

$⇒U=0.19\text{ J}$

At point Q total height is$R$

The potential energy at point$Q$is

$U=mgR$

$⇒U=0.032×9.80×12×{10}^{−2}$

$⇒U=0.038\text{ J}$

At top of the loop total height is

The potential energy at top of the loop is

$U=2mgR$

$⇒U=0.032×9.80×12×{10}^{−2}$

$⇒U=0.075\text{ J}$

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