You are here
-
๐ Revise, Reflect, Refine, Q11 (Page 138)
A 10.0 kg block is moving on horizontal floor with negligible friction. A variable force is applied on the block in its direction of motion from 0 m till 4 m (graph: force rises from 0 to 50 N over 0–1 m, stays 50 N from 1–3 m, falls back to 0 at 4 m). If the block had a kinetic energy of 180 J at 0 m, find the block's speed (i) at 0 m, and (ii) at 4 m. Does the block have negative acceleration in any portion of its motion?View answer
Answer-
(i) Speed at 0 m
- K = ½mv²
- 180 = ½ × 10 × v²
- v² = 36
- v = 6 m sโป¹
-
(ii) Speed at 4 m
- Work done = area under force-displacement graph
- Area = triangle (0–1 m) + rectangle (1–3 m) + triangle (3–4 m)
- Area = (½ × 1 × 50) + (2 × 50) + (½ × 1 × 50)
- Area = 25 + 100 + 25 = 150 J
- New kinetic energy = 180 + 150 = 330 J
- 330 = ½ × 10 × v²
- v² = 66
- v ≈ 8.1 m sโป¹
-
Negative acceleration?
- No. The force always acts in the direction of motion (it is never negative).
- So the block never has negative acceleration — it only speeds up or moves at constant speed.
-
(i) Speed at 0 m
Q11 - A 10.0 kg block is
Last updated at September 16, 2026 by Teachoo
