Anonymous
Anonymous asked in Science & MathematicsPhysics · 4 weeks ago

A sled of mass 1615 kg has four identical rockets. With all four rockets burning, the sled initial acceleration is 57.0 m/s2. If the rocket sled shown in the figure below starts with only one rocket burning, what is its acceleration? Assume that the force of friction opposing the motion is known to be 430 N. (Why is the acceleration not one-fourth of what it is with all rockets burning?)

Relevance

F = ma

If we let

F = force of one rocket

Ff = friction force

4F - Ff = ma

F = (ma + Ff) / 4

so the force for each rocket

F = (1615(57.0) + 430) / 4

F = ‭23,121.25‬ N

With only one rocket firing

a = F / m

a = (‭23,121.25‬ - 430) / 1615

a = 14.1 m/s²

acceleration with only one rocket is below one forth the maximum because a single rocket must still overcome ALL of the friction. With four rockets the friction force can be considered as divided equally among them.

I hope this helps

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• with 4 thrusters

a = (4F-Fr)/m

F = (m*a+Fr)/4 = (1615*57+430)/4 = 23,121 N

with a single thruster

a' = (F-Fr)/m = (23,121-430)/1615 = 14.05 m/sec^2

14.05*4 = 56.20 < 57

the reason is that friction remains the same and doesn't become a quarter of its value ; if Fr becomes Fr' = Fr/4 , then :

a'' = (F-Fr')/m = (23,121-430/4)/1615 = 14.25 m/sec^2 , which is just a quarter of 57; infact :

14.05*4 = 57