A block of mass 10 kg starting from rest is pulled up a smooth inclined plane making an angle of 30^ with the horizontal by a constant force parallel to the incline. If the power delivered by the pulling force at t = 4 s from the start is 560 W , the distance covered by the block in the first 4 s is _____ m . [Take g = 10 m s ⁻² ]

Step-by-step solution

Let the constant acceleration of the block be a and the pulling force be F . Using Newton's second law along the incline: F - mg 30^ = ma F = 10(a + 10 0.5) = 10(a + 5) The velocity of the block at t = 4 s is: v = u + at = 0 + a(4) = 4a The power delivered by the pulling force at t = 4 s is given by: P = F v 560 = 10(a + 5) 4a 560 = 40a(a + 5) a^2 + 5a - 14 = 0 (a + 7)(a - 2) = 0 Since acceleration must be positive in the direction of motion, a = 2 m s ⁻² . The distance covered by the block in the first 4 s is: s =