Asked by impulse and momentum
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.
Answers
Answered by
drwls
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
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
Answered by
Anonymous
1960J
Answered by
Shabnam P A
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.