Work Done by Forces While Pushing a Box
📘 Question:
Example: A person pushes a 20 kg box across 12 meters of carpet. The coefficient of kinetic friction is μₖ = 0.7. A force of 960 N is applied at an angle of 37° with respect to the horizontal.
Find:
- Work done by the applied force
- Work done by the friction force
- Work done by gravity
- Work done by the normal force
🔹 Step 1: Known Values
Mass (m) = 20 kg
Displacement (s) = 12 m
Force (F) = 960 N
Angle (θ) = 37°
μₖ = 0.7
g = 9.8 m/s²
🔹 Step 2: Resolve Applied Force
Horizontal component:
Fₓ = F × cos(θ) = 960 × cos(37°) ≈ 960 × 0.7986 = 766.69 N
Vertical component (downward):
Fᵧ = F × sin(37°) = 960 × 0.6018 = 577.74 N
Work by applied force (horizontal):
W₁ = Fₓ × s = 766.69 × 12 = 9200.28 J
🔹 Step 3: Work by Friction Force
Normal force on the box:
Fₙ = m × g + Fᵧ = (20 × 9.8) + 577.74 = 773.74 N
Frictional force:
fₖ = μₖ × Fₙ = 0.7 × 773.74 = 541.62 N
Work done by friction (opposite to motion):
W₂ = −fₖ × s = −541.62 × 12 = −6499.44 J
🔹 Step 4: Work by Gravity and Normal Force
Gravity acts downward, perpendicular to the horizontal displacement.
⇒ Work by gravity = 0 J
Normal force acts upward, also perpendicular to the displacement.
⇒ Work by normal force = 0 J
- Work by applied force: +9200.28 J
- Work by friction: −6499.44 J
- Work by gravity: 0 J
- Work by normal force: 0 J