Question

a bullet of mass 20g moving with a speed of 150m/s strikes a block and comes to rest after piercing 10cm into it. find the average force of resistance offered by the block

Answer 1

first we convert given parameters into stranded form n then solve them

Answer 2

Given :

Mass of bullet , m = 20 g = 0.02 kg .

Speed of bullet , u = 150 m/s .

Distance , covered before bullet stops , s = 10 cm = 0.1 m .

To Find :

The average force of resistance offered by the block .

Solution :

Acceleration is given by :

$v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{0^2-150^2}{2\times 0.1}\\\\a=-112500\ m/s^2$

So , the force is given by :

$F=0.02\times 112500\\\\F=2250\ N$

Therefore , the average force of resistance offered by the block is 2250 N .

Learn More :

Kinematics

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