When you drop a 0.35 kg apple, Earth exerts a force on it that accelerates it at

Question

When you drop a 0.35 kg apple, Earth exerts
a force on it that accelerates it at 9.8 m/s2
toward the earth’s surface. According to Newton’s third law, the apple must exert an equal but opposite force on Earth.
If the mass of the earth 5.98 × 10^24kg, what
is the magnitude of the earth’s acceleration
toward the apple?
Answer in units of m/s

A 59 kg boy and a 42 kg girl use an elastic rope
while engaged in a tug-of-war on a frictionless
icy surface.If the acceleration of the girl toward the
boy is 2.2 m/s2
determine the magnitude of
the acceleration of the boy toward the girl.
Answer in units of m/s2

Explanation / Answer

force on the apple by earth=m*g=.35*9.8
so force on earth by the apple(by magnitude)=(.35*9.8)(as per newtons third law),
so the acc of earth=force/(mass of earth)=(.35*9.8)/(5.98*10^24)=.574*10^(-24) m/s^2


by similar argument as per above 59*x=42*2.2
where x is the acceleration of the boy towards the girl.

which gives x=1.566m/s2

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