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yogui
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Help Spring Avarage acceleration and Spring Constant in a Spacecraft !
In NASA’s current concept, the lander has 4 legs, each with a plate on the end that will set firmly on the surface. Each leg has a spring to act as a shock absorber and each shock will compress 30 cm when the leg plates touch the surface at a velocity of 0.5 m/s. Assume the engines’ thrust goes instantly to zero at the time of contact with the surface and the lunar surface does not compress. What is the average acceleration of the vehicle after it touches the surface? What is the spring constant for each spring?
V = .5 m/s
M= 20,000 Kg
F=Kd Hookes law
d= 1/2(Vf+Vi)T
F=ma
What i would do is: F=Kd
F=ma
ma= K (.30 m)
20,000 ( ∆V/T)= K (.30)
∆V/T = (Vf-Vi )/t = ( 0 m/s - .5 m/s )/T
D= 1/2 (Vf+Vi) T
T= 2D/ (Vf+Vi)
T = 2 ( .30 m) / (.5 m/s)
T=1.2 s
replacing..
∆V/T = ( 0 m/s - .5 m/s )/ 1.2 s
= .416 m/s^2
20,000 ( ∆V/T)= K (.30)
20,000 ( .416)= K (.30)
8,333 N / .30 m = k
K= 27,777.77 N/m
it gave me a big number, I am not sure if my process is right because i have not taken this at my physics class yet, but based on reading that's what i get.
As for the average aceleration I am not sure.
i would say
a= ∆V/T = ( 0 m/s - .5 m/s )/ 1.2 s
a = .416 m/s^2
Any help is greatly appreciated, Thankz!
Homework Statement
In NASA’s current concept, the lander has 4 legs, each with a plate on the end that will set firmly on the surface. Each leg has a spring to act as a shock absorber and each shock will compress 30 cm when the leg plates touch the surface at a velocity of 0.5 m/s. Assume the engines’ thrust goes instantly to zero at the time of contact with the surface and the lunar surface does not compress. What is the average acceleration of the vehicle after it touches the surface? What is the spring constant for each spring?
V = .5 m/s
M= 20,000 Kg
Homework Equations
F=Kd Hookes law
d= 1/2(Vf+Vi)T
F=ma
The Attempt at a Solution
What i would do is: F=Kd
F=ma
ma= K (.30 m)
20,000 ( ∆V/T)= K (.30)
∆V/T = (Vf-Vi )/t = ( 0 m/s - .5 m/s )/T
D= 1/2 (Vf+Vi) T
T= 2D/ (Vf+Vi)
T = 2 ( .30 m) / (.5 m/s)
T=1.2 s
replacing..
∆V/T = ( 0 m/s - .5 m/s )/ 1.2 s
= .416 m/s^2
20,000 ( ∆V/T)= K (.30)
20,000 ( .416)= K (.30)
8,333 N / .30 m = k
K= 27,777.77 N/m
it gave me a big number, I am not sure if my process is right because i have not taken this at my physics class yet, but based on reading that's what i get.
As for the average aceleration I am not sure.
i would say
a= ∆V/T = ( 0 m/s - .5 m/s )/ 1.2 s
a = .416 m/s^2
Any help is greatly appreciated, Thankz!