# A bullet of mass 20g moving with a speed of 75m/s hits a fixed wooden plank and comes to rest after penetrating a distance of 5 cm.wht is the average resistive force exerted by wooden plank on bullet

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We don't know the average force by the bullet , so let it be m/s²

⇒

⇒

→ Therefore , we get acceleration as -56250 m/s²

⇒ Now , let us find the force :-

f = 0.02 kg × ( - 56250 )

f = -1125 Newton

Hence the average force is N

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

M=20 g=20/1000=.02 kg

v=0 m/s,since the bullet comes to rest

u=75 m/s

s=5 cm=5/100 m=.05 m

a=?

To find we can use v^2-u^=2as

0-75^2=2*a*.05

-5625=.1a

a=-5625/.1=-56250

to find the resistive force use this formulae

f=ma

f=0.02*-56250=-1125N

therefore the average resistive force is -1125 Newton

v=0 m/s,since the bullet comes to rest

u=75 m/s

s=5 cm=5/100 m=.05 m

a=?

To find we can use v^2-u^=2as

0-75^2=2*a*.05

-5625=.1a

a=-5625/.1=-56250

to find the resistive force use this formulae

f=ma

f=0.02*-56250=-1125N

therefore the average resistive force is -1125 Newton