The weight of man in lift moving in upward direction with an acceleration 'a' is 660 N. When the lift moves in the downward direction with the same acceleration, his weight is found to be 380 N. The real weight of the man when the lift is at rest is

Step-by-step solution

Let the mass of the man be m and the acceleration of the lift be a . When the lift moves upwards with acceleration a , the apparent weight is given by: W₁ = m(g + a) = 660 When the lift moves downwards with acceleration a , the apparent weight is given by: W₂ = m(g - a) = 380 Adding the two equations, we get: m(g + a) + m(g - a) = 660 + 380 2mg = 1040 mg = 520 N The real weight of the man when the lift is at rest is mg = 520 N. Answer: 520 N