SME
Grade 11 Science · Problem type
Hooke’s law: a spring’s stiffness
Use this for “A 2 kg mass hung on a spring stretches it 8 cm. What is k, and how much energy does the stretched spring store?”
Diagram
F = mg = 2 × 9.8 = 19.6 N
k = F/x = 19.6/0.08 = 245 N/m
U
=
1
2
kx²
=
1
2
×
245
×
0.08
2
U = 0.784 J
U is the triangle under the line.
This is an example. Type your own values to start a new problem.
m (kg)
×
9.8
m/s²
=
k (N/m)
×
x (m)
F
Force
calculated
U
Stored energy
calculated · J
Formulas
F
=
m
×
9.8
19.6 = 2 × 9.8
F
=
k
×
x
19.6 = 245 × 0.08
U
=
1
2
×
k
×
x
2
0.784
=
1
2
×
245
×
0.08
2
Assumptions
•
The spring pulls back in proportion to the stretch: F = kx, until it is overstretched.
•
A hung mass at rest stretches it until the spring’s pull equals the weight mg.
•
The stored energy is the area under the F–x line: ½kx².
Step-by-step
WE KNOW
m = 2 kg, x = 0.08 m
FIND
force (F), spring constant (k), stored energy (U)
Step 1 · Find force (F)
Use F = m × 9.8
At rest, the spring holds up the weight: 9.8 N for each kilogram.
F
=
2
×
9.8
F
=
19.6
N
Step 2 · Find spring constant (k)
Use F = k × x
Divide the force by the stretch.
k
=
F
x
k
=
19.6/0.08
k
=
245
N/
m
Step 3 · Find stored energy (U)
Use U = ½ × k × x²
The
triangle
under
the
F
–
x
line:
1
2
×
k
×
x
2
,
half
the
stretch
times
the
force
kx.
U
=
1
2
×
245
×
0.08
2
U
=
1
2
×
245
×
0.0064
U
=
122.5
×
0.0064
U
=
0.784
J
Check
Put the numbers back in. Both sides should match.
19.6 = 2 × 9.8 ✓
19.6 = 245 × 0.08 ✓
0.784
=
1
2
×
245
×
0.08
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.