# 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

2
by deeptiBera190

2015-09-26T11:42:37+05:30

### This Is a Certified Answer

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.
To solve the average  force exerted by the wooden plank on bullet use the given formula :-

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

⇒ Final velocity of bullet = v = 0

Initial velocity of the bullet = u = 75 m/s

Distance = s = 5cm =

⇒ Now , let us substitute the values in the formula :-

→ 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

2015-09-26T22:47:54+05:30

### This Is a Certified Answer

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