SME
Grade 8 Science · Problem type
Net force: a pull against friction
Use this for “A 50 N pull against 30 N of friction on an 8 kg box. What is the acceleration?” and “Why must you keep pushing to keep a box sliding at a steady speed?”
Diagram
Unbalanced: the box moves right.
m = 8 kg (mass)
a = 2.5 m/s² (acceleration)
This is an example. Type your own values to start a new problem.
n (N)
=
F (N)
−
f (N)
a (m/s²)
=
n (N)
÷
m (kg)
Formulas
n
=
F
−
f
(net
force
=
pull
−
friction)
20 = 50 − 30
a
=
n
÷
m
(acceleration
=
net
force
÷
mass)
2.5 = 20 ÷ 8
Assumptions
•
Two forces along a line: subtract the one pointing back from the one pointing forward.
•
Friction always points against the sliding.
•
Pull equal to friction: net force 0, so the box keeps its speed. That is why a steady slide still needs a steady pull.
Step-by-step
WE KNOW
F = 50 N (pull), f = 30 N (friction), m = 8 kg (mass)
FIND
net force (n), acceleration (a)
Step 1 · Find net force (n)
Use n = F − f (net force = pull − friction)
Friction points back, so take it away from the pull.
n
=
50
−
30
n
=
20
N
Step 2 · Find acceleration (a)
Use a = n ÷ m (acceleration = net force ÷ mass)
Only the net force speeds the box up: divide it by the mass.
a
=
20
8
a
=
2.5
m/
s
2
Check
Put the numbers back in. Both sides should match.
20 = 50 − 30 ✓
2.5 = 20 ÷ 8 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.