A 30 g bullet travelling initially at 500 ms-1 penetrates 12 cm into wooden block. The average force exerted will be

Question

A 30 g bullet travelling initially at 500 ms-1 penetrates 12 cm into wooden block. The average force exerted will be (A) 31250 N (B) 41250 N (C) 31750 N (D) 30450 N

Tardigrade Admin · Accepted Answer

From kinetic energy relation (1/2)mv2=F.s or F=(mv2/2s) (30 × 10-3 × (500)2/2 × 12× 10-2) 31250 N

Q. A 30 g bullet travelling initially at 500 $m s^{- 1}$ penetrates 12 cm into wooden block. The average force exerted will be

31250 N

41250 N

31750 N

30450 N

From kinetic energy relation
$\frac{1}{2} m v^{2} = F . s$
or $F = \frac{m v ^{2}}{2 s}$
$\frac{30 \times 1 0 ^{- 3} \times ( 500 ) ^{2}}{2 \times 12 \times 1 0 ^{- 2}}$
31250 N

Q. A 30 g bullet travelling initially at 500 ms−1 penetrates 12 cm into wooden block. The average force exerted will be