NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice
An observer and a vehicle, both start moving together from rest (towards the right) with acceleration 5 m s - 2 and 2 m s - 2 , respectively. There is a 2 kg block on the floor of the vehicle and coefficient of friction is μ = 0. 3 between their surface. Then the work done by the frictional force on the 2 kg block observed by the running observer, during the first 2 seconds of the motion
Options
- A24   J
- B-24 J
- C16   J
- D-16 J
Correct answer
B. -24 J
Step-by-step solution
f limiting = μ  mg = 0 . 3 × 2 × 1 0 = 6   N Force needed for acc. of 2   m   s - 2 F = m a = 4   N So friction force static, f = 4   N Displacement w.r.t. observer in t = 2 sec. is S = u rel t + 1 2 a rel t 2 = 0 + 1 2 - 3 × 4 = - 6   m W = f → . S → = 4 - 6 = - 2 4   J