JEE MainPhysicsLaws of Motion
A block of mass 5 kg is sliding at a speed of 5 m/s on a rough horizontal floor. The coefficient of kinetic friction between the block and the floor is 0.2 . A constant horizontal force is applied on the block in the direction of its motion. If the block covers a distance of 44 m in the next 4 s, the magnitude of the applied force is (Take g = 10 m/s ^2 ):
Options
- A15 N
- B25 N
- C5 N
- D37.5 N
Correct answer
B. 25 N
Step-by-step solution
From the second equation of motion, s = ut + 1 2 at^2 44 = 5(4) + 1 2 a(4)^2 44 = 20 + 8a a = 3 m/s ^2 The kinetic friction acting on the block is, f_k = _k mg = 0.2 5 10 = 10 N Applying Newton's second law in the horizontal direction, F - f_k = ma F - 10 = 5 3 F = 25 N Answer: 25 N