A body of mass 2kg is released from rest on a plane inclined at an angle of 60° to the horizontal. Calculate the acceleration of the body down the plane.[g ...

Answer Details

When a body of mass m is released on an inclined plane at an angle θ to the horizontal, the component of its weight acting down the plane is mgsinθ. The force acting parallel to the plane that causes the body to slide down is given by F = mgsinθ. The acceleration of the body down the plane is given by a = F/m = gsinθ. In this case, m = 2kg and θ = 60°. Therefore, the acceleration of the body down the plane is given by: a = gsinθ = 10ms\(^{-2}\) x sin(60°) = 8.7ms\(^{-2}\) Therefore, the correct answer is 8.7ms\(^{-2}\).