- #1
richievuong
- 35
- 0
A 1000kg plane is trying to make a forced landing on the deck of a 2000 kg barge at rest on the surface of a calm sea. The only frictional force to consider is between the plane's wheels and the deck, and this braking force is constant and equal to 1/4 of the plane's weight What must the minimum length of the barge be, in order that the plane can stop safely on the deck, if the plane touches down at the rear end of the deck with a velocity of 50m/s towards the front of the barge?
My work:
m1 = 1000kg
m2 = 2000kg
Ff = 1/4Fg
V1' = V2'
Fg = mg
=(1000)(9.81)
=9810N
Ff = 1/4Fg
= 1/4(9810)
=2452.5N
Fnet=Ff
P1(plane) = mv
= (1000)(50)
= 50000
P2(plane) = mv
= (1000)(0)
= 0
deltaP = (Fnet)t
50000 = 2452.5t
t = 20.387s
m1v1 + m2v2 = m1v1' + m2v2'
(1000)(50) + 2000(0) = 0 + (3000)v2'
50000 = 3000v2'
v2' = 16.667 m/s
a = V2-V1 / T
= (16.667-50) / 20.387
= -1.635m/s²
d = v1t + 1/2at²
= (50)(20.387) + 1/2(-1.635)(20.387)²
= 1019.35 - 339.777
=679.573m
Thats what i did but the answer is 340m...wondering what I did wrong...
My work:
m1 = 1000kg
m2 = 2000kg
Ff = 1/4Fg
V1' = V2'
Fg = mg
=(1000)(9.81)
=9810N
Ff = 1/4Fg
= 1/4(9810)
=2452.5N
Fnet=Ff
P1(plane) = mv
= (1000)(50)
= 50000
P2(plane) = mv
= (1000)(0)
= 0
deltaP = (Fnet)t
50000 = 2452.5t
t = 20.387s
m1v1 + m2v2 = m1v1' + m2v2'
(1000)(50) + 2000(0) = 0 + (3000)v2'
50000 = 3000v2'
v2' = 16.667 m/s
a = V2-V1 / T
= (16.667-50) / 20.387
= -1.635m/s²
d = v1t + 1/2at²
= (50)(20.387) + 1/2(-1.635)(20.387)²
= 1019.35 - 339.777
=679.573m
Thats what i did but the answer is 340m...wondering what I did wrong...