Dragging a load: work against friction, power, and the machine's efficiency

SPH3U Grade 11 Physics · Energy and Society

An electric winch drags a 500 kg sled of firewood 25.0 m across level frozen ground in 40.0 s, pulling horizontally on a cable. The coefficient of kinetic friction between the sled runners and the ground is 0.30, and the winch motor draws a steady 1.40 kW of electrical power the whole time. Find the friction force, the work done against it, the useful power delivered, the electrical energy consumed, and the efficiency of the winch.

Step 1 of 6 · solve for Weight

On level ground the surface has to hold up the entire weight, so mg is also the normal force — the one number friction is built from.

Rearranged for W
W=mgW = m g
Your values, in your units
W=(500 kg)gW = \left( 500\ \text{kg} \right) \, g
Answer
W=4.9W = 4.9

Carried onward at full precision, not this rounded figure.

Open the Weight (W = mg) solver →

Step 2 of 6 · solve for Kinetic friction force

Friction is what the cable is actually fighting. The sled is not speeding up, so the pull equals this force exactly — no more, no less.

Rearranged for fk
fk=μkNf_k = \mu_k N
4.9 kNcarried from step 1
Your values, in your units
fk=(0.3)(4,903.32 N)f_k = \left( 0.3 \right) \, \left( 4{,}903.32\ \text{N} \right)
Answer
fk=1.47f_k = 1.47

Carried onward at full precision, not this rounded figure.

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

Step 3 of 6 · solve for Work

Force times distance, with θ = 0° because the cable pulls along the motion. This is the useful work — every joule of it ends up as heat in the snow, which is exactly what the job required.

Rearranged for W
W=FdcosθW = F d \cos\theta
1.47 kNcarried from step 2
Your values, in your units
W=(1,471 N)(25 m)cos(0 )W = \left( 1{,}471\ \text{N} \right) \, \left( 25\ \text{m} \right) \, \cos \left( 0\ ^{\circ} \right)
Answer
W=36.8W = 36.8

Carried onward at full precision, not this rounded figure.

Open the Work (W = Fd cos θ) solver →

Step 4 of 6 · solve for Power

The same work spread over the 40.0 s haul. This is the winch's useful output, not what it draws from the battery.

Rearranged for P
P=WtP = \frac{W}{t}
36.8 kJcarried from step 3
Your values, in your units
P=(36,774.9 J)(40 s)P = \frac{\left( 36{,}774.9\ \text{J} \right)}{\left( 40\ \text{s} \right)}
Answer
P=919P = 919

Carried onward at full precision, not this rounded figure.

Open the Power (P = W/t) solver →

Step 5 of 6 · solve for Energy

Now the input side of the ledger. The motor is badged in kilowatts and the ledger is kept in joules, so the 1.40 kW becomes 1400 W before it meets the same 40.0 s.

Rearranged for E
E=PtE = P t
Your values, in your units
E=(1.4 kW)(40 s)E = \left( 1.4\ \text{kW} \right) \, \left( 40\ \text{s} \right)
Converted to base units
E=(1,400 W)(40 s)E = \left( 1{,}400\ \text{W} \right) \, \left( 40\ \text{s} \right)
Answer
E=56E = 56

Carried onward at full precision, not this rounded figure.

Open the Electrical Energy (E = Pt) solver →

Step 6 of 6 · solve for Efficiency

Efficiency is the ratio of the two energies, and this is the first step that reaches back to two different earlier answers at once. Compare it with 919.4 W ÷ 1400 W — identical, because both ran for the same 40.0 s.

Rearranged for η
η=WoutWin\eta = \frac{W_{out}}{W_{in}}
36.8 kJcarried from step 3
56 kJcarried from step 5
Your values, in your units
η=(36,774.9 J)(56,000 J)\eta = \frac{\left( 36{,}774.9\ \text{J} \right)}{\left( 56{,}000\ \text{J} \right)}
Answer
η=65.7\eta = 65.7

Carried onward at full precision, not this rounded figure.

Open the Machine Efficiency solver →

Why this order

An efficiency is never a single measurement; it is always a ratio of two independently computed energies, which is why this chain has to build both sides before it can say anything. The output side takes three steps — weight, friction, work — because the useful work is not something you read off a gauge: you have to know what force the job actually demanded. The input side takes one, because electrical energy is simply what the meter says. The classic error is to compare a force with a power, or the work done against friction with the motor's rated power, and the fix is dimensional: efficiency is joules over joules, so both sides must be energies.

Note that the 25.0 m and the 40.0 s make no difference to the answer. Drag the same sled twice as far in twice the time and both energies double, leaving η at 66%. That is the honest way to read the number: it says nothing about how hard the winch worked and everything about where the missing 34% went — gearbox friction, resistive heating in the motor windings, and the cable creaking over the drum. Efficiencies also multiply, so a 66% winch fed by a generator that is itself 30% efficient delivers about 20% of the fuel's energy to the firewood. That compounding is the whole reason the Energy and Society strand exists: a slogan about conservation becomes an argument only once it is a chain of numbers.

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