All Solutions
Page 101: Section Review
Known:
$$
begin{align*}
m&=10 mathrm{kg} \
g_{Earth}&=9.8 mathrm{m/s^2} \
g_{Moon}&=1.6 mathrm{m/s^2}
end{align*}
$$
Unknown:
$$
begin{align*}
F_{g(Earth)}/F_{g(Moon)}&=?
end{align*}
$$
$$
begin{align*}
F_g&=mg \
\
F_{g(Earth)}/F_{g(Moon)}&=dfrac{mg_{Earth}}{mg_{Moon}} \
&=dfrac{g_{Earth}}{g_{Moon}} \
&=dfrac{9.8 mathrm{m/s^2}}{1.6 mathrm{m/s^2}} \
&=boxed{6.125}
end{align*}
$$
From the force ratio we can see that the force of gravity on the Moon is approximately 6 times smaller than the force of gravity on Earth.
Known:
$$
begin{align*}
F&=9 mathrm{N} \
m&=65 mathrm{kg}
end{align*}
$$
Unknown:
$$
begin{align*}
a&=?
end{align*}
$$
$$
begin{align*}
F&=mcdot a \
a&=dfrac{F}{m}
end{align*}
$$
begin{align*}
a&=dfrac{F}{m} \
&=dfrac{9 mathrm{N}}{65 mathrm{kg}} \
&=boxed{0.1385 mathrm{m/s^2}}
end{align*}
$$
a=0.1385 mathrm{m/s^2}
$$
Known:
$$
begin{align*}
m&=1 mathrm{kg} \
F_s&=9.3 mathrm{N} \
g&=9.8 mathrm{s^2}
end{align*}
$$
$$
begin{align*}
F_g&=mcdot g \
&=1 mathrm{kg} cdot 9.8 mathrm{m/s^2} \
&=9.8 mathrm{N}
end{align*}
$$
$$
begin{align*}
F&=mcdot a \
F_s-F_g&=mcdot a \
\
a&=dfrac{F_s-F_g}{m} \
&=dfrac{9.3 mathrm{N}-9.8 mathrm{N}}{1 mathrm{kg}} \
&=boxed{-0.5 mathrm{m/s^2}}
end{align*}
$$
Known:
$$
begin{align*}
F_{Marcos}&=22 mathrm{N} \
F_{cat}&=-19.5 mathrm{N} \
a&=6.25 mathrm{m/s^2}
end{align*}
$$
Unknown:
$$
begin{align*}
m&=?
end{align*}
$$
$$
begin{align*}
F_{net}&=F_{Marcus}+F_{cat} \
&=22 mathrm{N}-19.5 mathrm{N} \
&=2.5 mathrm{N}
end{align*}
$$
$$
begin{align*}
F&=ma\
m&=dfrac{F}{a} \
&=dfrac{2.5 mathrm{N}}{6.25 mathrm{m/s^2}} \
&=boxed{0.4 mathrm{kg}}
end{align*}
$$
Since the person is falling at a constant speed, we can assume that, until the parachute opens, there is no force on the diver (As then the acceleration is $0$ in Newton’s second law).
Of course, in the real world, a person could never go in a free fall and have a constant velocity, due to Earth’s gravitational force that is always giving the acceleration of $a=-g$. This is just an assumption for this specific problem, where we could approximate the velocity to be constant for a short period of the free fall.
When the parachute opens, there is a drag force due to air molecules hitting the big surface of the parachute. These molecules are technically pushing the parachute up, as from the person’s relative perspective the molecules are hitting the person in the free fall.
This drag force is thus pushing upwards. So the drag force gives upwards acceleration. This creates the effect of initial constant velocity lowering. So if we take the downwards direction to be positive, the acceleration is negative (Downwards deceleration).
