Toboggan on the flat: energy in, friction out, distance and time

SPH3U Grade 11 Physics · Energy and Society

A toboggan and rider, 52.0 kg together, shoot off the bottom of a packed run onto a long flat stretch of snow at 5.40 m/s. The coefficient of kinetic friction between the waxed base and the snow is 0.12, and the pair coasts in a straight line until friction brings them to rest.

Given

m = 52 kg — Toboggan and rider together
v₀ = 5.4 m/s — Speed entering the flat
μₖ = 0.12 — — Kinetic friction, wax on snow

Determine

(a)the kinetic energy at the bottom of the run
(b)the friction force acting on the toboggan
(c)the net work friction does, and the stopping distance
(d)the time the toboggan takes to stop

Step 1 of 6(a) · solve for Kinetic energy

The energy budget for the entire runout is fixed at this instant: 758 J and not a joule more. Everything friction takes, it takes from here.

Rearranged for Eₖ

E_k = \tfrac{1}{2} m v^{2}

m

v

Your values, in your units

E_k = \tfrac{1}{2} \, \left( 52\ \text{kg} \right) \, \left( 5.4\ \text{m/s} \right)^{2}

Answer

E_k = 758.16\ \text{J}

in

Carried onward at full precision, not this rounded figure.

Open the Kinetic Energy solver →

Step 2 of 6(b) · solve for Weight

On level snow the surface holds up the whole weight, so mg doubles as the normal force. That is a level-ground privilege — on the slope the toboggan just left, N was mg cos θ and this shortcut would have been wrong.

Rearranged for W

W = m g

m

Your values, in your units

W = \left( 52\ \text{kg} \right) \, \left( 9.80665\ \text{m/s}^{2} \right)

Answer

W = 509.95\ \text{N}

in

Carried onward at full precision, not this rounded figure.

Open the Weight (W = mg) solver →

Step 3 of 6 · solve for Kinetic friction force

Friction is built on the normal force, never directly on the mass: f = μₖN = 61.2 N. Note what is NOT in the formula — speed. Kinetic friction drags just as hard at 5.4 m/s as in the last crawling centimetre.

Rearranged for fk

f_k = \mu_k N

μk

N509.95 Ncarried from step 2

Your values, in your units

f_k = \left( 0.12 \right) \, \left( 509.946\ \text{N} \right)

Answer

f_k = 61.193\ \text{N}

in

Carried onward at full precision, not this rounded figure.

Open the Kinetic Friction Force (f = μₖN) solver →

Step 4 of 6(c) · solve for Net work

The work-energy theorem states the ending first: to go from 5.40 m/s to rest, the net work on the toboggan must be −758 J — negative, because energy is leaving. Writing +758 here is the sign error that later "proves" friction speeds sleds up.

Rearranged for W

W = \tfrac{1}{2} m \left(v^{2} - v_0^{2}\right)

m

v

v0

Your values, in your units

W = \tfrac{1}{2} \, \left( 52\ \text{kg} \right) \left(\left( 0\ \text{m/s} \right)^{2} - \left( 5.4\ \text{m/s} \right)^{2}\right)

Answer

W = -758.16\ \text{J}

in

Carried onward at full precision, not this rounded figure.

Open the Work–Energy Theorem solver →

Step 5 of 6 · solve for Displacement

Friction points backwards along the motion, so θ = 180° and cos θ = −1: a negative work divided by a negative force-projection gives a positive 12.4 m. The two minus signs are not pedantry — they are the direction bookkeeping that makes the distance come out forwards.

Rearranged for d

d = \frac{W}{F \cos\theta}

W-758.16 Jcarried from step 4

F61.193 Ncarried from step 3

θ

Your values, in your units

d = \frac{\left( -758.16\ \text{J} \right)}{\left( 61.1935\ \text{N} \right) \, \cos \left( 180\ ^{\circ} \right)}

Answer

d = 12.39\ \text{m}

in

Carried onward at full precision, not this rounded figure.

Open the Work (W = Fd cos θ) solver →

Step 6 of 6(d) · solve for Time

Constant friction means constant deceleration, so the average speed over the stop is (5.40 + 0)/2 = 2.70 m/s and t = d divided by THAT — 4.59 s, exactly twice the 12.4/5.4 a hasty division would give.

Rearranged for t

t = \frac{2 d}{v_0 + v}

d12.39 mcarried from step 5

v0

v

Your values, in your units

t = \frac{2 \, \left( 12.3896\ \text{m} \right)}{\left( 5.4\ \text{m/s} \right) + \left( 0\ \text{m/s} \right)}

Answer

t = 4.5887\ \text{s}

in

Carried onward at full precision, not this rounded figure.

Open the Displacement from Average Velocity solver →

Why this order

This is the work-energy theorem doing a job the kinematic equations could also do, and doing it more honestly. The energy route needs no acceleration and no time: the budget (758 J) is set in part (a), the spending rate per metre (61.2 N — a joule is a newton-metre) in part (b), and the stopping distance is literally budget over rate. The chain routes that division through W = Fd cos θ so the signs stay visible: the theorem in part (c) says the net work must be −758 J, friction at θ = 180° supplies negative work, and the distance emerges positive because the two negatives belong to each other. Students who drop the sign on either half get a negative stopping distance — an answer that is not slightly wrong but is telling you exactly which convention was broken.

Two structural facts deserve notice. First, the mass never mattered: KE = ½mv² and f·d = μmg·d both scale with m, so d = v₀²/(2μg) — a 100 kg pair on the same wax stops in the same 12.4 m, which is why stopping distance is a property of the SURFACE pairing and the speed, the same fact behind every highway-code braking table. Second, the v₀² is the safety lesson: half again as fast off the hill, 5.4 → 8.1 m/s, means 2.25 times the runout, 28 m, and a sledding hill that suddenly needs a much longer flat before the fence. And the 758 J did not vanish — it melted a microscopic film of snow along the track, which is also the reason a toboggan barely slides at −30 °C when the snow is too cold to lubricate: the same μ that set this whole problem is itself temperature's creature.

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