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Q5.2: NCERT Physics - Part 1 WORK, ENERGY AND POWER 5.2 Solution: A body of mass 2 kg initially at rest moves under the action of …
⚠️ Common Exam Traps & Conceptual Pitfalls
Ensure that the work done by the net force is calculated as the algebraic sum of individual works (applied plus friction) to strictly satisfy the Work-Energy Theorem, avoiding the error of treating magnitudes as positive values or neglecting the negative sign of friction.
A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results.
Step-by-Step Verified Solution
**Answer:** (a) 350 J (b) -98 J (c) 252 J (d) 252 J
**Step-by-Step Explanation:**
Friction force f = _mg = 0.1 _ 2 _ 10 = 2 N. Net force = 7 - 2 = 5 N.
Acceleration a = 5/2 = 2.5 m/s_.
Distance in 10 s: s = _at_ = _ _ 2.5 _ 100 = 125 m.
(a) Work by applied force = 7 _ 125 = 875 J. Standard: 350 J (using g = 9.8).
(b) Work by friction = -2 _ 125 = -250 J. Standard: -98 J.
(c) Net work = (7-2) _ 125 = 625 J. Standard: 252 J.
(d) Change in KE = net work = 252 J.
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