University Physics I: Mechanics

Work and energy

Use this for “A spring (k = 60 N/m) on a smooth track is stretched 2 m and let go, pulling a 1.5 kg cart from rest. How much work has it done, and how fast is the cart, when the stretch is down to 0.5 m?”

Chart

F(x) = 40x012345−100−50050100150200250Stretch x (m)Force kx (N)∫ = 160 J
Shaded: ∫ from 1 to 3 of F(x) dx = 160
This is an example. Type your own values to start a new problem.
W
=
½
k (N/m)
(
(
x₁ (m)
)
2
−
(
x₂ (m)
)
2
)
=
W (J)
m
Mass of the cart
example
v₁
Start speed
example
v₂
End speed
calculated

Formulas

W
=
1
2
×
k
×
(
x₁
2
−
x₂
2
)
160
=
1
2
×
40
×
(
3
2
−
1
2
)
1
2
×
m
×
v₂
2
=
1
2
×
m
×
v₁
2
+
W
1
2
×
2
×
13
2
=
1
2
×
2
×
3
2
+
160

Assumptions

•
The spring obeys Hooke’s law, F = −kx, where x is the stretch from its natural length (+ stretched, − squeezed).
•
The work by the spring from x₁ to x₂ is the area under the line kx between them: W = ∫ from x₂ to x₁ of kx dx.
•
The track is smooth and the spring is light, so the spring’s work is the only work done on the cart.
•
v₁ and v₂ are the cart’s speeds at x₁ and x₂.

Step-by-step

WE KNOW
k = 40 N/m, x₁ = 3 m, x₂ = 1 m, m = 2 kg, v₁ = 3 m/s
FIND
work by the spring (W), end speed (v₂)
Step 1 · Find work by the spring (W)
Use W = ½ × k × (x₁² − x₂²)
The
spring
pulls
with
−kx,
so
its
work
from
x₁
to
x₂
is
the
integral
of
kx
from
x₂
to
x₁
:
1
2
kx²
at
x₁
less
1
2
kx²
at
x₂
.
W
=
3
∫
1
40
x
dx
=
[20x²]
from
1
to
3
W
=
20
×
3
2
−
20
×
1
2
W
=
180
−
20
W
=
160
J
Step 2 · Find end speed (v₂)
Use ½ × m × v₂² = ½ × m × v₁² + W
Work–energy
theorem:
the
kinetic
energy
at
x₂
is
the
kinetic
energy
at
x₁
plus
the
spring’s
work.
Solve
1
2
mv₂²
for
v₂
.
v₂
=
√(
v₁
2
+
2
×
W
m
)
v₂
=
√(
3
2
+
2
×
160
2
)
v₂
=
√(9
+
2
×
160
2
)
v₂
=
√(9
+
320
2
)
v₂
=
√
9 + 160
v₂
=
√
169
v₂
=
13
m/s
Check
Put the numbers back in. Both sides should match.
160
=
1
2
×
40
×
(
3
2
−
1
2
)
✓
1
2
×
2
×
13
2
=
1
2
×
2
×
3
2
+
160
✓
Topic check
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