Step 1 — Resolve forces on the incline
Break the weight into components relative to the plane:
• Component of gravitational force parallel to the plane:
Fgravity,‖ = mg sin θ. With m = 1 kg, g = 10 m/s² and sin 60° = √3/2:
Fgravity,‖ = 1 × 10 × (√3/2) = 5√3 N ≈ 8.660 N.
• Component of weight perpendicular to the plane (normal force magnitude when no other vertical forces are present):
N = mg cos θ. With cos 60° = 1/2:
N = 1 × 10 × (1/2) = 5 N.
Step 2 — Compute the frictional force
Friction (kinetic or limiting static, as applicable) is:
Ffric = μN.
Substitute μ = 0.1 and N = 5 N:
Ffric = 0.1 × 5 = 0.5 N.
Note: the frictional force is constant here because μ and N are constant along the plane.
Step 3 — Work done against friction over 10 m
Work done against a constant force equals force × distance (with force component in the direction opposite to motion):
Wfric = Ffric × distance.
Thus
Wfric = 0.5 N × 10 m = 5 J.
Step 4 — (Optional) Energy perspective & total work required
If you also want the work needed to raise the block against gravity (useful if calculating the push force required at constant speed):
Height gained along the incline: h = distance × sin θ = 10 × (√3/2) = 5√3 m ≈ 8.660 m.
Increase in potential energy (work against gravity):
ΔPE = m g h = 1 × 10 × 5√3 = 50√3 J ≈ 86.602 J.
Therefore, if the block is moved up at constant velocity (no net kinetic energy change), the total work done by the applied push is
Wtotal = ΔPE + Wfric ≈ 86.602 + 5 = 91.602 J.
(But the question specifically asked only for work done against friction — that is 5 J.)
Step 5 — Units, sign convention & interpretation
• We report work done against friction as a positive quantity (energy expended to overcome friction) — here 5 J. • If asked for work done by friction (the force itself), that would be negative (−5 J) because friction acts opposite to displacement. • If you are computing the work done by the pushing force at constant speed, include both ΔPE and Wfric as shown above.
Common mistakes to avoid
• Using g = 9.8 vs 10: here we used g = 10 m/s² as given; with g = 9.8 answers change slightly. • Forgetting that normal force = mg cos θ, not mg. • Forgetting that friction = μN (and that μ multiplies the normal force, not mg directly). • Omitting the factor sin θ when finding vertical height from incline distance.
Final answer
Work done against friction = 5 J.
