General Principles And Processes Of Isolation Of Elements

{Angular momentum formula}
Thus, the correct answer is 14.4 kg m s .

Step 1 — Resolve forces on the incline

Break the weight into components relative to the plane:

• Component of gravitational force parallel to the plane:
F_gravity,‖ = mg sin θ. With m = 1 kg, g = 10 m/s² and sin 60° = √3/2:
F_gravity,‖ = 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:

F_fric = μN.

Substitute μ = 0.1 and N = 5 N:

F_fric = 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):

W_fric = F_fric × distance.

Thus

W_fric = 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

W_total = ΔPE + W_fric ≈ 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 W_fric 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.