If the acceleration of the girl toward the boy is 3.0 m/s2, find the magnitude of the acceleration of the boy toward the girl.
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the force the boy exerts on the girl is equal in magnitude and opposite in direction to the force of the girl on the boy
if the girl accelerates at 3m/s/s, the force on her is
F = m a = 41kg x 3 m/s/s = 123N
this is the same force that acts on the boy, whose mass is 69kg, so we have
F = m a
a (boy) = 123N/ 69 kg = 1.78m/s/s
if the girl accelerates at 3m/s/s, the force on her is
F = m a = 41kg x 3 m/s/s = 123N
this is the same force that acts on the boy, whose mass is 69kg, so we have
F = m a
a (boy) = 123N/ 69 kg = 1.78m/s/s
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What we can assume is that the force on each person is the same.
F=ma
for the girl a=3 and m=41
so the force acting on her must be F= 41(3) or 123 N
The same force acting on the 69kg boy would create acceleration a of
a = F/m or 123/69 or 1.78 m/s/s
F=ma
for the girl a=3 and m=41
so the force acting on her must be F= 41(3) or 123 N
The same force acting on the 69kg boy would create acceleration a of
a = F/m or 123/69 or 1.78 m/s/s
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If the girl is accelerating at 3m/s^2, the force on her must be F = ma = 41*3 = 123N
This has to be the force on the boy too since they are connected by the rope
His acceleration would then be a = F/m = 123/69 = 1.78m/s^2
This has to be the force on the boy too since they are connected by the rope
His acceleration would then be a = F/m = 123/69 = 1.78m/s^2
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(41 x 3)/66 = 1.864m/sec.
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