Worked examples — once it moves
The same 60 N block, now sliding. The dynamic friction between it and the table is 15 N. Three more examples — and this is where chapter 4 comes back in.
Example 4 — find the acceleration
The problem
The block is pulled horizontally with a force of 25 N and is sliding along the table. Find its acceleration. (Take g = 10 m s−2.)
Solution
It is moving, so the friction acting is the dynamic value, 15 N.
unbalanced force = applied force − friction
F = 25 − 15 = 10 N
The mass is not given, so get it from the weight:
m = weight ÷ g = 60 ÷ 10 = 6 kg
Now the second law:
F = m a
10 = 6 × a
a = 10 ÷ 6 = 1.7 m s−2 (to 2 s.f.)
Two traps here. The 25 N is not the F in F = m a — only the unbalanced 10 N is. And 60 is a weight in newtons, not a mass in kilograms; F = m a needs the mass.
Example 5 — sliding steadily
The problem
The pull is eased back until it is exactly 15 N, the block still sliding. What happens to the block?
Solution
unbalanced force = 15 − 15 = 0
F = m a, so a = 0 ÷ 6 = 0
No acceleration, and the block is already moving, so it slides on at a constant velocity.
Newton's first law, seen in the wild. Balanced forces do not mean "at rest" — they mean "no change in the motion", and here the motion that does not change is a steady slide.
Example 6 — add a load
The problem
The block is brought to rest and a 20 N block is placed on top of it. What is the new normal reaction, and what happens to the force needed to start it moving?
Solution
R = total weight = 60 + 20
R = 80 N
Limiting friction increases with R, so the force needed to start it is now more than 18 N.
The surfaces in contact are the same wood and the same table, and the contact area has not changed either — so the only reason anything changed is R.
Remember the method
Once it slides: unbalanced force = applied − dynamic friction, then F = m a. Check whether you have been given a mass or a weight before substituting, and check your units at the end — kilograms and newtons, never the other way round.