A bullet of mass 50 g travelling it 500 m/s penetrates 100 cm into a wooden block. The average force exerted on the block is

Question

A bullet of mass 50 g travelling it 500 m/s penetrates 100 cm into a wooden block. The average force exerted on the block is (A)  6.25× 105N  (B)  26.5× 104N  (C)  6.25× 103N  (D)  2.65× 104N

Tardigrade Admin · Accepted Answer

From the relation Fs=(1/2)mv2-(1/2)mu2 ∴ -F(1)=(1/2)( (50/1000) )[(0)2-(500)2] -F=-(25/1000)× 500× 500 F=6.25× 103N

Q. A bullet of mass 50 g travelling it 500 m/s penetrates 100 cm into a wooden block. The average force exerted on the block is

$6.25 \times 10^{5} N$

$26.5 \times 10^{4} N$

$6.25 \times 10^{3} N$

$2.65 \times 10^{4} N$

From the relation $F s = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$ $∴$ $- F (1) = \frac{1}{2} (\frac{50}{1000}) [(0)^{2} - (500)^{2}]$ $- F = - \frac{25}{1000} \times 500 \times 500$ $F = 6.25 \times 10^{3} N$

Q. A bullet of mass 50 g travelling it 500 m/s penetrates 100 cm into a wooden block. The average force exerted on the block is

6.25×105N

26.5×104N

6.25×103N

2.65×104N

From the relation Fs=21​mv2−21​mu2 ∴ −F(1)=21​(100050​)[(0)2−(500)2] −F=−100025​×500×500 F=6.25×103N

$6.25 \times 10^{5} N$

$26.5 \times 10^{4} N$

$6.25 \times 10^{3} N$

$2.65 \times 10^{4} N$

From the relation $F s = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$ $∴$ $- F (1) = \frac{1}{2} (\frac{50}{1000}) [(0)^{2} - (500)^{2}]$ $- F = - \frac{25}{1000} \times 500 \times 500$ $F = 6.25 \times 10^{3} N$