SME
University Physics I: Mechanics
Newton's laws
Use this for “A 4 kg block on a table (μₖ = 0.25) is pulled by a 2 kg block hanging over a pulley at the edge. Find the acceleration and the string’s tension.”
Diagram
a
=
(m₂
−
μₖm₁)
g
m₁ + m₂
a
=
(2
−
0.25
×
4)
×
9.81
4 + 2
a
=
1.635
m/
s
2
Block 1: m₁a = T − μₖm₁g = 16.35 − 9.81 = 6.54 N
Block 2: m₂a = m₂g − T = 19.62 − 16.35 = 3.27 N
A light pulley: the tension is the same on both sides.
This is an example. Type your own values to start a new problem.
a (m/s²)
=
m₂ (kg)
−
×
m₁ (kg)
+
×
9.81
Add the two blocks’ laws, m₂g − T = m₂a and T − μₖm₁g = m₁a: the tension cancels, and dividing by m₁ + m₂ gives a.
T
Tension
worked out (not typed)
Formulas
a
=
(
m₂
−
μₖ
×
m₁
)
×
9.81
m₁
+
m₂
1.635
=
(2
−
0.25
×
4)
×
9.81
4 + 2
T
=
m₂
×
(9.81
−
a
)
16.35 = 2 × (9.81 − 1.635)
Assumptions
•
A light string over a light, frictionless pulley, so the tension T is the same on both sides; g = 9.81 m/s².
•
The string doesn’t stretch, so both blocks move together with one acceleration a.
•
The table block is already sliding, so kinetic friction μₖm₁g holds it back.
•
If m₂ ≤ μₖm₁, friction is enough to hold the blocks: once at rest, nothing slides.
Step-by-step
WE KNOW
m₁ = 4 kg, m₂ = 2 kg, μₖ = 0.25
FIND
acceleration (a), tension (T)
Step 1 · Find acceleration (a)
Use a = (m₂ − μₖ × m₁) × 9.81 ÷ (m₁ + m₂)
Add
m₂g
−
T
=
m₂
a
and
T
−
μₖm₁g
=
m₁
a
:
T
cancels.
The
hanging
weight
less
the
friction
moves
both
masses.
a
=
(2
−
0.25
×
4)
×
9.81
4 + 2
a
=
(2
−
1)
×
9.81
6
a
=
9.81
6
a
=
1.635
m/
s
2
Step 2 · Find tension (T)
Use T = m₂ × (9.81 − a)
Newton’s
second
law
on
the
hanging
block:
m₂g
−
T
=
m₂
a
,
so
T
is
m₂g
less
m₂
a
.
T
=
2
×
(9.81
−
1.635)
T
=
2
×
8.175
T
=
16.35
N
Check
Put the numbers back in. Both sides should match.
1.635
=
(2
−
0.25
×
4)
×
9.81
4 + 2
✓
16.35 = 2 × (9.81 − 1.635) ✓
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A projectile’s position and velocity at a time
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An Atwood machine: two hanging blocks
Topic check
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