A free-body diagram isolates an object and shows all external forces. It is the starting map for force problems.
Block on table — arrows for weight (down), normal (up), friction (left), applied push (right).
Weight \(W\) acts vertically downward due to gravity.
The table pushes upward with an equal normal reaction \(N\).
Static friction \(f\) lies along the surface, resisting any push.
Practice: Drag each force name to its arrow on the block, then hit “Check Answers” to self-check your labels.
Arrow 1
Arrow 2
Arrow 3
Arrow 4
Start with weight; it always points straight down from the block.
When forces stop balancing, \(F_{\text{net}}\) appears and the block’s speed starts to change.
Use \(a = F_{\text{net}}/m\). Bigger net force or smaller mass means faster acceleration.
1 N applied to 1 kg produces 1 m s⁻², confirming Newton’s Second Law.
We push a 2 kg block with 10 N. Kinetic friction is 2 N. Find the acceleration.
Forward push 10 N, backward kinetic friction 2 N. Choose forward as positive.
\(F_{\text{net}} = 10\,\text{N} - 2\,\text{N} = 8\,\text{N}\) forward.
\(a = \dfrac{F_{\text{net}}}{m} = \dfrac{8}{2} = 4\ \text{m/s}^2\) forward.
Always subtract opposing friction before using \(F = ma\).
A 5 kg block has 15 N rightward and 5 N leftward forces. What is its acceleration?
First find net force, then apply \(a = F_{\\text{net}}/m\).
Net force = 10 N rightward, so \(a = 10/5 = 2\;\\text{m/s}^2\) to the right.
Remember: subtract opposing forces, then divide by the mass.
Begin by sketching a clear free-body diagram of the block.
On a level plane, weight \(mg\) downward balances normal force \(N\) upward.
Static or kinetic friction acts opposite the intended or actual motion.
Net horizontal force \( \Sigma F_x = ma \); its direction sets the block’s acceleration.
Thank You!
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