Question

A ball of mass 500 grams is suspended with the help of a light, inextensible string of length 100 cm. The maximum value of to which the string rotates from the mean position A to reach the extreme position C is ${60}^{\circ }$. Consider $g=10m/s^2$

i. What will be the potential energy of the ball when it reaches half the maximum height from position A?

[1 mark]

• A. P.E. = 2.25 J
• B. P.E. = 0.5 J
• C. P.E. = 1.25 J
• D. P.E. = 0.75 J

Answer 1

The correct option is C P.E. = 1.25 J

Given,
Mass of the ball = 0.5 kg
Length of the string = 100 cm = 1 m



$In\Delta OCD,\Rightarrow OC=1mand\cos {60}^o=\frac{OD}{OC}=\frac{OD}{1}\Rightarrow OD=0.5m$

Now, AD = OA - OD = 1 - 0.5 = 0.5 m

The length AD represents the maximum height that the ball can achieve while going from the mean position A to the extreme position C.

Total energy at the extreme position C = K.E. + P.E.
= 0 + (m$\times$g$\times$AD)

When the ball reaches half the maximum height, its vertical distance from A will be$\frac{0.5}{2}=0.25m$

The potential energy at F, ${P.E.}_F=m\times g\times AE=0.5\times 10\times 0.25=1.25J$