Question

A bullet of mass 0.025. kg is moving with a velocity of 200 m/s is stopped within 5 cm by a target. The average force exerted by the bullet is _ KN. Pls answer fast. ( Who ever write the correct solution for the given answer quickly will be marked as the brainliest answer.. hurry up.It is urgent) Hint- the answer is 10

Answer 1

Explanation:

Given

m = 0.025kg

u = 200 m/s

s = 5cm = 0.05m

v = 0 m/s ( Because the bullet stops)

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

${0}^{2} = {200}^{2} + 2a(0.05)$

$a = \frac { - 200 \times 200}{2 \times 0.05} = - 400000 \: m {s}^{ - 2}$

As we know,

F = ma

F = 0.025 x (-400000) = -10000 N = -10KN

F = -10kN

Answer 2

Answer

Average retarding force exerted by bullet is 10 k-N

Given

A bullet of mass 0.025. kg is moving with a velocity of 200 m/s is stopped within 5 cm by a target

To Find

Force exerted by bullet

Solution

Given values are ,

$\rm Mass,m=0.025\ Kg\\\\\rm Initial\ velocity,u=200\ m/s\\\\\rm Distance,s=5\ cm=0.05\ m\\\\\rm Final\ velocity,v=0\ m/s\ [Finally\ stops]\\\\\rm Force,F=?\ N$

Apply 3 rd equation of motion ,

$\implies \rm v^2-u^2=2as\\\\\implies \rm (0)^2-(200)^2=2a(0.05)\\\\\implies \rm 0-40000=0.1a\\\\\implies \rm a=-400000\ m/s^2$

Apply 2nd law of Newton sir's motion ,

$\implies \rm F=ma\\\\\implies \rm F=(0.025)(-400000)\\\\\implies \rm F=\dfrac{25}{1000}\times -400000\\\\\implies \rm F=-25\times 400\\\\\implies \rm F=-10000\ N\\\\\implies \rm F=-10\ kN$

Note : Negative sign ( - ve ) of force denotes retarding force

So , Average retarding force exerted by the bullet is 10 k-N .

More Info

SI unit of Force is Newton

SI unit of Mass is Kg

SI unit of acceleration is m/s²

SI unit of velocity is m/s

SI unit of distance is meter