Q. A steel ball of mass 5 g is thrown downward with velocity $10\ {ms}^{- 1}$ from height 19.5 m. It penetrates sand by 50 cm. The change in mechanical energy will be $(g = 10{ms}^{- 2})$
A) 1 J
B) 1.25 J
C) 1.5 J
D) 1.75 J
Explanation
(b) Given m= 5g = 0.005kg, h=19.5m,
$x = 50cm = 0.5m,\ \ \ v = 10ms^{- 1},\ g = 10ms^{- 2}$
The change in mechanical energy
$\mathrm{\Delta}U = mg(h + x) + \frac{1}{2}mv^{2}$
$= 0.005 \times 10(19.5 + 0.5) + \frac{1}{2} \times 0.005 \times (10)^{2}$
=$0.005 \times 10 \times 20 + \frac{1}{2} \times 0.005 \times 100$
$= 1 + 0.25 = 1.25j$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.