If the rocket launches with a force of 10N, what is the force on the launch pad?
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Answer:
<h2>1/7 kg</h2>Explanation:
Find the diagram attached for better understanding of the question.
Given the mass of one of the blocks to be 1.0kg and accelerates downward at 3/4g.
g = acceleration due to gravity.
Let the block accelerating downward be M, mass of the other body be 'm' and the acceleration of the body M be 'a'.
M = 1.0 kg and a = 3.4g
According to newton's second law;
For body of mass m;
T - mg = ma ... (1)
For body of mass M;
Mg - T = Ma ... (2)
Adding equation 1 ad 2;
+Mg -mg = ma + Ma
Ma-Mg = -mg-ma
M(a-g) = -m(a+g)
Substituting M = 1.0 kg and a = 3/4g into the resulting equation;
3/4 g-g = -m(3/4 g+g)
3/4 g-g = -m(7/4 g)
-g/4 = -m(7/4 g)
1/4 = 7m/4
28m = 4
m = 1/7 kg
Therefore the mass of the other box is 1/7 kg
here we can say that net force on the student vertically upwards will be counter balance by his weight downwards
Let net force F is exerted by each hand
so here we will have
so the force exerted by each hand will be 414.1 N