SME
Calculus II · Problem type
Work stretching a spring
Use this for “A force of 30 N holds a spring stretched 0.15 m past its natural length. How much work stretches it from 0.15 m to 0.4 m?”
Chart
Shaded: ∫ from 0.1 to 0.3 of F(x) dx = 8
This is an example. Type your own values to start a new problem.
F
=
k (N/m)
×
s (m)
=
F (N)
W
=
∫
k
x
dx
=
W (J)
Formulas
F
=
k
×
s
20 = 200 × 0.1
W
=
k
2
×
(
x₂
2
−
x₁
2
)
8
=
200
2
×
(
0.3
2
−
0.1
2
)
Assumptions
•
Hooke’s law: stretched x past its natural length, the spring pulls back with force F = kx, where k is its spring constant.
•
The force changes as the spring stretches, so the work is an integral: W = ∫ from x₁ to x₂ of kx dx = ½k(x₂² − x₁²).
•
Every stretch x is measured from the natural length, not from where the spring starts.
•
Force is in N, stretch in m and work in J (1 J = 1 N·m).
Step-by-step
WE KNOW
F = 20 N, s = 0.1 m, x₁ = 0.1 m, x₂ = 0.3 m
FIND
spring constant (k), work done stretching the spring (W)
Step 1 · Find spring constant (k)
Use F = k × s
Hooke’
s
law
turned
around:
divide
the
force
by
the
stretch
it
holds.
k
=
F
s
k
=
20
0.1
k
=
200
N/m
Step 2 · Find work done stretching the spring (W)
Use W = k ÷ 2 × (x₂² − x₁²)
An
antiderivative
of
kx
is
kx²
2
.
Take
its
value
at
x₁
from
its
value
at
x₂
.
W
=
200
2
×
(
0.3
2
−
0.1
2
)
0.3
∫
0.1
200
x
dx
=
[100x²]
from
0.1
to
0.3
=
8
W
=
8
J
Check
Put the numbers back in. Both sides should match.
20 = 200 × 0.1 ✓
8
=
200
2
×
(
0.3
2
−
0.1
2
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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