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Answer:
Part (i) the initial acceleration of the rocket is 6.98 m/s²
Part(ii) the floor pushes on the power supply at 120m altitude by a force of 31.68 N
Explanation:
Part (i) the initial acceleration of the rocket.
For the rocket to accelerate, the force applied to it must overcome gravitational force due to its own weight.
[tex]F_{Net} = M(a+g)\\\\1930 = 115(a+9.8)\\\\a +9.8 =\frac{1930}{115} \\\\a +9.8 = 16.78\\\\a = 16.78-9.8\\\\a = 6.98 \frac{m}{s^2}[/tex]
Part(ii) how hard the floor pushes on the power supply at 120 m altitude
At 120 m height, the acceleration of the rocket is 6.98 m/s², which is the same as the power supply.
given force on power supply;
F = 18.5 N
Applying Newton's second law of motion, the mass of the power supply = 18.5/9.8
= 1.888 kg
The force on power supply at this altitude = m(a+g)
= 1.888(6.98 +9.8)
= 1.888(16.78)
= 31.68 N
Therefore, the floor pushes on the power supply at 120 m altitude by a force of 31.68 N
The initial acceleration of the rocket is 6.98 m/s².
The floor pushes on the power supply at 120m altitude by a force of 31.68N.
Based on the information given, the initial acceleration will be calculated thus:
F = M (a + g)
1930 = 115(a + 9.8)
1930 = 115a + 1127
115a = 1930 - 1127
115a = 803
a = 803 / 115
a = 6.986.98 m/s².
The second law of motion will be used to solve the remaining question. This will be:
F = m(a + g)
F = 1.888(6.98 + 9.8)
F = 1.888 × 16.78
F = 31.68N
Therefore, the floor pushes on the power supply at 120m altitude by a force of 31.68N.
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