Answer key: Force & Friction Worksheet
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Goal: Find out when friction holds a block still, when it starts to slide, and how net force, mass and acceleration are related (F = ma).
Completed results table
| Setting | Friction | Net force | Acceleration | State |
|---|---|---|---|---|
| Default (5 kg, 20 N, μ = 0.3) | 14.715 N | 5.285 N | 1.057 m/s² | Sliding |
| Smooth surface (μ = 0) | 0 N | 20 N | 4 m/s² | Sliding |
| Held by friction (10 N, μ = 0.5) | 10 N | 0 N | 0 m/s² | At rest |
| Sliding (60 N, μ = 0.3) | 14.715 N | 45.285 N | 9.057 m/s² | Sliding |
| 14 N push | 14 N | 0 N | 0 m/s² | At rest |
| 15 N push | 14.715 N | 0.285 N | 0.057 m/s² | Sliding |
Answers
With “Held by friction” the block does not move, even though you push with 10 N. How big is the friction force, and why?
10 N, exactly equal to the push. Static friction matches the applied force up to its maximum, μN = 0.5 × 5 × 9.81 ≈ 24.5 N, so the net force is 0 and the block stays at rest.
Between 14 N and 15 N the block changes from resting to sliding. Use the maximum friction μN to explain where the switch happens.
Maximum static friction is μN = 0.3 × 5 × 9.81 = 14.715 N. A 14 N push is less than that, so friction cancels it. A 15 N push exceeds it, so the block slides, but the small net force of 0.285 N gives an acceleration of only 0.057 m/s².
Compare the default (μ = 0.3) with “Smooth surface” (μ = 0). How does removing friction change the net force and the acceleration?
With no friction the whole 20 N is the net force, so a = F ÷ m = 20 ÷ 5 = 4 m/s². With μ = 0.3, friction takes 14.715 N away, leaving 5.285 N and an acceleration of 1.057 m/s².
Use your “Sliding” row to check Newton’s second law: does net force ÷ mass equal the acceleration?
Yes: 45.285 N ÷ 5 kg = 9.057 m/s², which matches the acceleration readout.
Predict, then test: if you double the mass to 10 kg and still push with 60 N, will the acceleration be half of the “Sliding” value?
No, it is less than half: 3.057 m/s² instead of about 4.53 m/s². Doubling the mass also doubles the normal force, so friction doubles to 29.43 N, leaving a net force of only 30.57 N.
Challenge: this model uses one friction coefficient for both resting and sliding. In real life the kinetic coefficient is usually smaller. What would you notice when pushing a heavy box?
It takes a bigger push to get the box moving than to keep it moving. Once it starts sliding, friction drops, so the same push suddenly produces a larger net force and the box lurches forward.