Please find attachment. Read this well and use it to enhance as well as develop your understanding and knowledge about the given assignment. If you have any queries feel free to send me a message. Thank you very much! BEST OF LUCK :) regards,Dr. Charles Kennedy (expertsolution) solution: Drawing F.B.D of crate and applying equation of motion we can solve this problem as below- Comment

The acceleration of the crate a = ( v^2 - u^2)/2S = (8^2 - 0^2) / 2*8 = 4 ms^-2 The kinetic friction Fk = μ * N = 0.6*N Σ Fy = m*ay N + P sin 40° - 43*9.81 = 43*0 N = 421.83 - 0.64*P Σ Fx = m*ax P cos 40° - Fk = m*a P * 0.77 - 0.6*N = 43*4 Plugging value of N 0.77*P - 0.6(421.83 - 0.64*P) = 43*4 0.77*P - 253.1 + 0.38*P = 172 P = 369.65 So, the magnitude of force P acting on the crate is 369.65