Question

A steel ball of mass $5 \,g$ is thrown downward with velocity $10 \,m/s$ from height $19.5\, m$. It penetrates sand by $50\, cm$. The change in mechanical energy will be : $(g = 10\, m/s )$

• A. 1 J
• B. 1.25 J
• C. 1.51 J
• D. 1.75J

Answer 1

Answer: (B)

The change in mechanical energy
$ \Delta U =mg (h+x) +\frac{1}{2}mv^2 $
Here $m = 5\,g = 0.005 \,kg. h = 19.5\, m$
$x = 50\, cm = 0.5\, m, v = 10 \,ms^{-1} $
hence $ \Delta U = 0.005 \times\, 10919.5 +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=1.25\, J $