A block of mass 5 kg rests on a smooth horizontal surface. It is pulled by a light string that makes an angle of 37^ above the horizontal with a constant force of 50 N . Simultaneously, a constant horizontal opposing force of 10 N acts on the block. The acceleration of the block is (Take 37^ = 3 5 , 37^ = 4 5 and g = 10 m s ⁻² )

Step-by-step solution

The horizontal component of the pulling force is F_ x = F 37^ = 50 4 5 = 40 N . The vertical component of the pulling force is F_ y = F 37^ = 50 3 5 = 30 N . The weight of the block is mg = 5 10 = 50 N . Since F_ y The net force acting on the block in the horizontal direction is F_ net = F_ x - f_ opposing . Substituting the given values, F_ net = 40 - 10 = 30 N . Using Newton's second law, the acceleration of the block is a = F_ net m = 30 5 = 6 m s ⁻² . Answer: 6 m s ⁻²