An aircraft weighing 6400 pounds lands at a speed of 10 ft/Sec and stops in 10 Seconds. What force was generated by the brakets (asuming gravity as 32 ft/sec).
- -400 Lbs.
- -200 Lbs.
- -640 Lbs.
Explanation from PART66Online
The aircraft mass is weight / g = 6400 / 32 = 200 slugs. Decelerating from 10 ft/s to rest in 10 s gives a = -1 ft/s squared, so braking force = m x a = 200 x (-1) = -200 lb. The negative sign shows the force opposes the motion.
R
Rajaram12 asked · 22 Jun 2013
Plz solve this
Community Comments (3)
P
PART66Online
24 Jun 2013
F = m * a
Where:
F - Force
m - Mass
a - Acceleration
m = 6400 / 32 = 200
a = 10 / 10 = -1 (negative because deceleration)
F = 200 * (-1) = 200
Where:
F - Force
m - Mass
a - Acceleration
m = 6400 / 32 = 200
a = 10 / 10 = -1 (negative because deceleration)
F = 200 * (-1) = 200
R
ravi.0813
2 Dec 2014
Actually unit of weight is Newton(N) not pound. So aircraft weight should be 6400 Newtons.
J
Jordi
14 Jun 2020
yeah i agree with Ravi i think this question shoud change answers because when we say weight we mean the force which is given by W=m*g and mass which is found by the W=m*g.Can you please change it.
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