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Spring dart launcher: stiffness, stored energy, launch speed

SPH4U Grade 12 Physics · Energy and Momentum

A student clamps a toy dart launcher to a bench and investigates its spring. A force gauge shows that compressing the plunger 6.0 cm takes a steady 24.0 N. The launcher is then loaded with a 60.0 g foam dart, compressed the same 6.0 cm, aimed straight up, and fired; friction in the barrel is negligible.

Given
  • x = 6 cm — Compression of the plunger
  • F = 24 N — Force at full compression
  • m = 60 g — Mass of the dart
Determine
  1. (a)the spring constant of the launcher spring
  2. (b)the elastic energy stored at full compression
  3. (c)the speed of the dart as it leaves the barrel
  4. (d)the height the dart can climb above the muzzle
Step 1 of 4(a) · solve for Spring constant

A spring constant is newtons per METRE, so the 6.0 cm goes in as 0.060 m and k comes out at 400 N/m. Divide by 6 instead of 0.060 and you get a limp 4 N/m spring that could not launch a feather.

Rearranged for k
k=Fxk = \frac{F}{x}
Your values, in your units
k=(24 N)(6 cm)k = \frac{\left( 24\ \text{N} \right)}{\left( 6\ \text{cm} \right)}
Converted to base units
k=(24 N)(0.06 m)k = \frac{\left( 24\ \text{N} \right)}{\left( 0.06\ \text{m} \right)}
Answer
k=400 N/mk = 400\ \text{N/m}

Carried onward at full precision, not this rounded figure.

Open the Hooke's Law solver →

Step 2 of 4(b) · solve for Elastic potential energy

The energy is ½kx², not F·x. The force ramps from zero at rest length up to 24.0 N at full compression, so the average force over the squeeze is half the peak — that is where the ½ comes from, and skipping it doubles every answer downstream.

Rearranged for U
U=12kx2U = \tfrac{1}{2} k x^{2}
400 N/mcarried from step 1
Your values, in your units
U=12 (400 N/m) (6 cm)2U = \tfrac{1}{2} \, \left( 400\ \text{N/m} \right) \, \left( 6\ \text{cm} \right)^{2}
Converted to base units
U=12 (400 N/m) (0.06 m)2U = \tfrac{1}{2} \, \left( 400\ \text{N/m} \right) \, \left( 0.06\ \text{m} \right)^{2}
Answer
U=720 mJU = 720\ \text{mJ}

Carried onward at full precision, not this rounded figure.

Open the Elastic Potential Energy solver →

Step 3 of 4(c) · solve for Speed

With no friction, every stored joule rides out on the dart: Eₖ = 0.72 J. The dart is weighed in grams and ½mv² wants kilograms, so 60.0 g becomes 0.0600 kg before the square root.

Rearranged for v
v=2Ekmv = \sqrt{\tfrac{2 E_k}{m}}
720 mJcarried from step 2
Your values, in your units
v=2 (0.72 J)(60 g)v = \sqrt{\tfrac{2 \, \left( 0.72\ \text{J} \right)}{\left( 60\ \text{g} \right)}}
Converted to base units
v=2 (0.72 J)(0.06 kg)v = \sqrt{\tfrac{2 \, \left( 0.72\ \text{J} \right)}{\left( 0.06\ \text{kg} \right)}}
Answer
v=4.899 m/sv = 4.899\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Kinetic Energy solver →

Step 4 of 4(d) · solve for Height

The same 0.72 J spent against gravity instead: h = U/mg. Notice the answer must agree with h = v²/2g built from step 3 — one energy, two currencies, and the exchange rate is fixed.

Rearranged for h
h=Umgh = \frac{U}{m g}
720 mJcarried from step 2
Your values, in your units
h=(0.72 J)(60 g) (9.80665 m/s2)h = \frac{\left( 0.72\ \text{J} \right)}{\left( 60\ \text{g} \right) \, \left( 9.80665\ \text{m/s}^{2} \right)}
Converted to base units
h=(0.72 J)(0.06 kg) (9.80665 m/s2)h = \frac{\left( 0.72\ \text{J} \right)}{\left( 0.06\ \text{kg} \right) \, \left( 9.80665\ \text{m/s}^{2} \right)}
Answer
h=1.2237 mh = 1.2237\ \text{m}

Carried onward at full precision, not this rounded figure.

Open the Gravitational Potential Energy (U = mgh) solver →

Therefore the spring's stiffness is 400 N/m, it stores 0.72 J when fully compressed, and the dart leaves the barrel at 4.90 m/s — enough to climb about 1.22 m above the muzzle before gravity wins.

Why this order

This chain runs the entire energy ledger of a spring toy, and the order is the order in which the quantities become measurable. Stiffness comes first because it takes only a ruler and a force gauge — no motion at all. Stored energy needs that stiffness. Launch speed needs the stored energy. Height needs either. The step that costs the most marks is (b): F·x = 1.44 J is off by exactly a factor of two, and it is a seductive error because for a constant force W = Fd is correct. A spring's force is not constant — it grows linearly from zero — so the work is the area of a triangle, not a rectangle, and ½ · 24.0 N · 0.060 m says the same thing as ½kx².

Parts (c) and (d) are two withdrawals from the same 0.72 J account, and comparing them is the habit worth building. The speed route gives v = √(2U/m) = 4.90 m/s; the height route gives h = U/mg = 1.22 m; and v²/2g reproduces the height exactly, because both are the same energy read in different units. Notice what the dart's mass does and does not touch: a heavier dart leaves slower and tops out lower, but the STORED energy in part (b) is a property of the spring alone — which is why one launcher design ships with darts of many masses, all obeying the same 0.72 J budget. That budget idea, energy fixed at the source and spent along the path, is the same reasoning used to size everything from crossbow prods to aircraft-carrier catapults.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.