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Physics HW Help?
The thrust produced by a single jet engine creates a force of F = 91000 N. It takes the jet (with a mass of m = 9900 kg) a distance of d = 0.75 km to take off.
A.) What is the take-off speed of the jet vt in m/s?
B.) How far in meters would you need to depress a giant spring k = 100,000 N/m in order to launch the jet at the same speed without help from the engine?
3 Answers
- AshLv 75 days agoFavorite Answer
A)
v² = u² + 2as
v² = u² + 2(F/m)s
v² = 0² + 2[(91000N)/(9900 kg)](750 m)
v = 117 m/s ← take off speed of jet
B)
PE of spring = KE of jet
½kx² = ½mv²
x² = mv²/k
x² = (9900 kg)(117 m/s)²/(10000 N/m)
x = 36.8 m ← distance by which the spring needs to be depressed
- oubaasLv 75 days ago
V = √2*a*d = √2*750*91/9.9 = 117.4 m/sec
k*x^2 = m*V^2
x = V√m/k = 117.4*√9.9/100 = 117.4/10*√9.9 = 36.9 m
- billrussell42Lv 75 days ago
a = F/m = 91000/9900 = 9.192 m/s²
v² = v₀² + 2ad
v² = 0² + 2(9.192)(750)
v = 117.4 m/s
KE = ½mV² = KE = ½(9900)(117.4)² = 68250000 J
Springs and weightk = F/d k is spring constant in N/m F is force and d is deflectionPE = ½kd²
k = 100,000 N/m
PE = ½kd² = 68250000
d² = 2•68250000 / 100000 = 2•682.5
d = 36.9 m
