Question

A bullet of 25 gm with a speed 200 m/s stops at a distance of 5 cm. Find the resistive force ?( please help me) ​

Answer 1

Answer:

$10 {}^{4} newton$

Explanation:

here,

v=0 m/s

u=200m/s

m=25×(10) ^-3 kg

$v {}^{2} = u {}^{2} + 2as$

$0 {}^{2} = 200 {}^{2} + 2a \times 0.05$

$a = - 4 \times 10 {}^{5}$

F = ma

F = 25×(10)^-3 × -4 × (10)^5

$= - 10 {}^{4}$

since the resistive force is (10)^4 N