A block of mass m is placed on a frictionless inclined plane making an angle of 30^ with the horizontal. A constant external force of 21 ~N is applied on the block directed up the incline. If the block accelerates down the incline at 2 ~m ~s ⁻² , what is the value of m ? (Take g = 10 ~m ~s ⁻² )

Step-by-step solution

Let the mass of the block be m . The forces acting on the block along the inclined plane are: 1. The component of its weight acting down the incline: m g 30^ 2. The external force acting up the incline: F = 21 ~N Since the block accelerates down the incline at a = 2 ~m ~s ⁻² , the net force is directed downwards along the incline. According to Newton's second law: m g 30^ - F = m a Substituting the given values ( g = 10 ~m ~s ⁻² , 30^ = 0.5 ): m(10)(0.5) - 21 = m(2) 5m - 21 = 2m 3m = 21 m = 7 ~kg Note: If the block