NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice
A block of mass m is stationary with respect to the wedge of mass M moving with uniform speed v on horizontal surface. Work done by friction force on the block in t seconds is
Options
- Azero
- B– m g v t 2 sin 2 θ
- C– m g v t 2
- D– m g v t 2 sin θ
Correct answer
B. – m g v t 2 sin 2 θ
Step-by-step solution
Since block doesn't move with respect towedge, so friction force should balance the component of weight along the inclined plane. f = m g sin θ Component of displacement along the inclined plane in time t is given by v t cos θ Work done by f = – m g sin θ cos θ v t = – m g sin 2 θ 2 v × t