Question

A 69.0-kg astronaut is floating in space, luckily he has his trusty 28.0-kg physics book. He throws his physics book and accelerates at 0.0130 m/s2 in the opposite direction. What is the magnitude of the acceleration of the physics book?

Answer 1

Answer:

0.032 $m/s^2$

Explanation:

Given :

Weight of the astronaut = 69 kg

Weight of the physics book = 28 kg

Acceleration of the astronaut = 0.0130 $m/s^2$

The force that is applied on the astronaut :

$F=ma$

   $=69 \times 0.013$

   = 0.897 N

Therefore, by Newton's 3rd law, we know that the force applied on the physics book is also F = 0.897 N

Therefore, the acceleration of the physics book is given by :

$a = \frac{\text{Force on physics book}}{\text{mass of physics book}}$

$a = \frac{0.897}{28}$

a = 0.032 $m/s^2$

Hence, the acceleration of the physics book is  0.032 $m/s^2$.

Answer 2

Answer:

The acceleration of astronaut is 5.27 x 10^-3 m/s^2.

Explanation:

mass of astronaut, M = 69 kg

Mass of book, m = 28 kg

acceleration of book, a = 0.013 m/s^2

Let the acceleration of astronaut is A.

According to the Newton's third law, for every action there is an equal and opposite reaction.

So, the force acting on the book is same as the force acting on the astronaut but the direction is opposite to each other.

M A = m a

69 x A = 28 x 0.013

A = 5.27 x 10^-3 m/s^2