4.15 A small 7.80- k g rocket burns fuel that exerts a time-varying upward force

Question

4.15

A small 7.80-kg rocket burns fuel that exerts a time-varying upward force on the rocket. This force obeys the equation F=A+Bt2. Measurements show that at t=0, the force is 139.0N , and at the end of the first 2.00s , it is 155.0N .

Find the constant A.

Find the constants B.

Find the net force on this rocket the instant after the fuel ignites.

Find the acceleration of this rocket the instant after the fuel ignites.

Find the net force on this rocket 5.00s after fuel ignition.

Find the acceleration of this rocket 5.00s after fuel ignition.

Suppose you were using this rocket in outer space, far from all gravity. What would its acceleration be 5.00s after fuel ignition?

Explanation / Answer

F = Bt^2 + A at t=0, f=139N
at t=2.0, f=155N

{rearrange for A}

A = F - Bt^2 so at t=0 F = 139

{sub 0 for t and 139 for F to find A}

therefore: A = 139 - B(0)^2 = 139

now

F = Bt^2 + 139 therefore 155 = Bt^2 + 139
{ finally rearrange to solve for B}

B = (155-139) / (2)^2 =4

the net force the moment the fuel ignites?

the mass of the rocket is 7.8 kg, therefore the force due to gravity at sea level is 9.8ms/s*7.8 kg = 76.44N and the upwards force on the rocket at t=0, we already know is 139N
so net force = 139 - 76.44 N = 62.56 N

acceleration = net force / mass
therefore a = 62.56 /7.8 kg =8.02 m/s^-2

{use original formula}
(e) F = 4(5^2) + 139 = 239N
so net = (239-76.44)N = 162.56 N

(f) again, a = f/m therefore a = 162.56/7.8 kg
therefore a = 20.84 ms^-2

(g) in space we can neglect gravitational force, and we have:

F = 239 N , do a = 239 / 7.8 = 30.64 m/s^-2

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