Average resistance (force) * (penetration distance) = (kinetic energy at impact)
F = M g H/X = 40*50/1 = 2000 N
Avg velocity during penetration *(Time) = (Penetration distance)
Time = X/(15.7 m/s) = 0.064 s
a body weighing 40N drops freely from a hieght of 50m and penetrates into the ground by 100cm.find the average resistance to penetration and the time of penetration.
3 answers
1960J
Change in K E = work done
1/2m(v2^2-v1^2) =FS
Here v1 is the initial velocity and v2 ia the final velocity
V1=√2gh= √2×9.8×50=31.3 m/s( dropped from cliff)
V2=P*x= 0×1=0(pentration velocity)
1/2*(40/9.8)*(0-31.3^2) =mgx-Px
-1999. 36=40*9.8*1-P*1
P=2039.88 N
1/2m(v2^2-v1^2) =FS
Here v1 is the initial velocity and v2 ia the final velocity
V1=√2gh= √2×9.8×50=31.3 m/s( dropped from cliff)
V2=P*x= 0×1=0(pentration velocity)
1/2*(40/9.8)*(0-31.3^2) =mgx-Px
-1999. 36=40*9.8*1-P*1
P=2039.88 N
1 answer