SME
University Physics I: Mechanics · Problem type
Work by a force F = cxⁿ
Use this for “A force F = 0.5x³ N pushes a cart along a track from x = 2 m to x = 4 m. How much work does it do?”
Chart
Shaded: ∫ from 1 to 4 of F(x) dx = 126
This is an example. Type your own values to start a new problem.
∫
c
x
dx
=
W (J)
Formulas
W
=
c
×
x₂
n
+ 1
−
x₁
n
+ 1
n
+ 1
126
=
6
×
4
2 + 1
−
1
2 + 1
2 + 1
Assumptions
•
The force acts along x and depends only on where the object is: F = cxⁿ, with c in N/mⁿ so that F is in newtons.
•
The work from x₁ to x₂ is the area under F between them: W = ∫ from x₁ to x₂ of cxⁿ dx.
•
x is at least 0, so xⁿ is defined for every power n from 0 to 4; n = 0 is a constant force.
Step-by-step
WE KNOW
c = 6, n = 2, x₁ = 1 m, x₂ = 4 m
FIND
work by the force (W)
Step 1 · Find work by the force (W)
Use W = c × (x₂ⁿ⁺¹ − x₁ⁿ⁺¹) ÷ (n + 1)
Power
rule:
an
antiderivative
of
cxⁿ
is
cxⁿ⁺¹
÷
(
n
+
1).
Take
its
value
at
x₂
less
its
value
at
x₁
.
W
=
4
∫
1
6
x
² dx
=
[2x³]
from
1
to
4
W
=
2
×
4
3
−
2
×
1
3
W
=
128
−
2
W
=
126
J
Check
Put the numbers back in. Both sides should match.
126
=
6
×
4
2 + 1
−
1
2 + 1
2 + 1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Work and energy
Main lesson
Force and equilibrium from U(x) = ax³ − bx²
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Energy down a rough slope: heat and the end speed
Use this for “A 40 kg sled starts down a 50 m slope at 20° moving at 3 m/s, with μₖ = 0.08. How much heat does friction make, and how fast is the sled at the bottom?”