A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of \( 60^\circ \) by a force of 10 N parallel to the inclined surface. When the block is pushed up by 10 m along the inclined surface, the work done against frictional force is:
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.
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Approach Solution -2
Step 1: {Calculate normal force}
The normal force \( N \) on the inclined plane is:
\[
N = mg \cos 60^\circ
\]
Step 2: {Determine frictional force}
\[
F_f = \mu N = \mu mg \cos 60^\circ
\]
Substituting values:
\[
F_f = (0.1) (1) (10) \left( \frac{1}{2} \right) = 0.5 { N}
\]
Step 3: {Calculate work done against friction}
\[
W = F_f \times d
\]
\[
= 0.5 \times 10 = 5 { J}
\]
Thus, the correct answer is (B) 5 J.
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