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 )

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

Tardigrade Admin · Accepted Answer

The change in mechanical energy Δ U =mg (h+x) +(1/2)mv2 Here m = 5 g = 0.005 kg. h = 19.5 m x = 50 cm = 0.5 m, v = 10 ms-1 hence Δ U = 0.005 × 10919.5 +0.5 +(1/2)× 0.005 × (10)2 =0.005 × 10 × 20 +(1/2)× 0.005× 100=1.25 J

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

1 J

1.25 J

1.51 J

1.75J

The change in mechanical energy ΔU=mg(h+x)+21​mv2 Here m=5g=0.005kg.h=19.5m x=50cm=0.5m,v=10ms−1 hence ΔU=0.005×10919.5+0.5+21​×0.005×(10)2 =0.005×10×20+21​×0.005×100=1.25J

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