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How do you draw the free-body diagram and determine the net force for each scenario: (1) a book at rest on a tabletop, (2) a hockey puck sliding across ice after being struck, (3) an egg in free fall with no air resistance, (4) a book pushed rightward across a desk with friction, and (5) a skateboard colliding with a wall and rebounding?

Answer
  1. Book at rest: forces are weight $mg$ down and normal force $N$ up, so $F_\text{net}=0$. 2) Sliding puck: $mg$ down, $N$ up, and kinetic friction $f_k$ opposite the motion, so the net force is horizontal opposite the motion with magnitude $f_k$ (vertical net is 0). 3) Free-falling egg (no air resistance): only $mg$ downward, so $F_\text{net}=mg$ downward. 4) Book pushed right: $mg$ down, $N$ up, applied force $F_\text{app}$ right, friction $f$ left, so $F_\text{net}=F_\text{app}-f$ to the right (vertical net is 0). 5) Skateboard at collision: $mg$ down, ground normal up, and a wall normal force pushing away from the wall, so the net force is mainly away from the wall at that instant.
Explanation

What you are doing in each free-body diagram

A free-body diagram (FBD) is a dot (the object) with arrows for every external force acting on it. Then add forces as vectors to find the net force $\vec F_\text{net}$.

1) Book at rest on a tabletop

Forces on the book

  • Weight: $mg$ downward.
  • Normal force from table: $N$ upward.

Net force Since the book is at rest (no acceleration), the forces balance: $$N = mg \quad\Rightarrow\quad \vec F_\text{net}=\vec 0.$$

2) Hockey puck sliding across the ice

Assume the puck is moving to the right.

Forces on the puck

  • Weight: $mg$ downward.
  • Normal force from ice: $N$ upward.
  • Kinetic friction: $f_k$ to the left (opposite the motion).

Net force Vertical forces cancel: $N = mg$ (no vertical acceleration). Horizontal net force is friction: $$\vec F_\text{net} = -f_k\,\hat{i}$$ So the puck slows down.

3) Egg in free fall (neglect air resistance)

Forces on the egg

  • Only gravity: $mg$ downward.

Net force $$\vec F_\text{net} = mg \text{ (downward)}$$ This gives acceleration $a=g$ downward.

4) Book pushed rightward across a desk with friction

Assume motion is to the right.

Forces on the book

  • Weight: $mg$ downward.
  • Normal force: $N$ upward.
  • Applied force: $F_\text{app}$ to the right.
  • Friction: $f$ to the left (kinetic friction $f_k$ if sliding; static friction $f_s$ if not yet moving).

Net force Vertical net is zero: $N=mg$. Horizontal net: $$F_\text{net, x} = F_\text{app} - f$$

  • If $F_\text{app} > f$, net force is rightward and the book accelerates right.
  • If $F_\text{app} = f$, net force is 0 and the book moves at constant speed (or stays at rest).

5) Skateboard colliding with a wall and rebounding

Assume the skateboard is moving right and hits a wall in front of it.

Forces at the instant of collision

  • Weight: $mg$ downward.
  • Normal force from ground: $N_\text{ground}$ upward.
  • Normal force from wall: $N_\text{wall}$ to the left (the wall pushes the board away from itself).
  • (Optional, depending on the level of detail): friction at the ground or wall could act, but the key rebound force is $N_\text{wall}$.

Net force At that moment, the largest unbalanced force is usually horizontal from the wall:

  • Vertical: typically $N_\text{ground} \approx mg$ so vertical net is about 0.
  • Horizontal: net force is to the left, approximately $$\vec F_\text{net} \approx -N_\text{wall}\,\hat{i}$$ This leftward net force changes the skateboard’s velocity and can make it rebound.
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Skills You Achive
free-body-diagrams newtons-laws vector-addition friction-forces net-force

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