An open knife of mass m is dropped from a height h on a wooden floor. if the blade penetrates up to the depth d into the wood,average resistance offered by the wood to the knife is?????

							

knife is dropped from A with u= 0 m/s
FROM A----->B
when it strikes the ground at B its velocity is:
v2 - u2 = 2as
v2- 0 = 2 (g) (h)
v= √2gh----------------------------------------(1)
after striking the ground it travels a distance ''d''
in this case u= √2gh {from (1)}
v=0
s= d
therefore, v2- u2= 2as
              0 - 2gh= 2(a)(d)
            -2gh/2d=a
        a= -gh/d
Now, force applied by the ground is = ma= m(-gh/d) = -mgh/d.
						

							
On conserving mechanical energy we get velocity before penetration i.e{½mu²+mgh=½mv²}we got (v=√2gh)
Now u=√2gh
V=0
Distance=s travelled by blade during its penetration.using eqñ (2as=v²-u²)we get acceleration to be (gh/s).now drawing it free body diagram we get an eqñ(mg-N=ma)putting value of a i.e acceleration we N=mg[1+h/s].where N is normal reaction or resistive force by the wooden surface during its penetration.
						

							
The initial KE  + final kinetic energy+ (+ve) WD +(-ve) WD will be equal to final Work done.....
KE1+(W+)+(W-)=KEf
0+mg(h+d)-Fr(d)=0
mg(h+d)=Fr(d)
Fr=mg(1+h/d)