Question

elevator is moving up with an acceleration g

5

. The apparent weight of a 60 kg man standing in the lift

is [Take g = 10 ms­​

Answer 1

When he is moving upwards  

F_net = m * (a + g)

= 60 * (4.9 + 9.8)

= 882 N

His Weight will be 882 N

Answer 2

let at any time t, q = EC (1 – e^-t/CR) E = Energy stored = q^2/2c = E^2C^2/2c (1 – e^-t/CR)^2 = E^2C/2 (1 – e^-t/CR)^2 R = rate of energy stored = dE/dt = -E^2C/2 (-1/RC)^2 (1 – e^-t/RC)e^-t/RC = E^2/CR . e^-t/RC (1 – e^-t/CR) dR/dt = E^2/2R [-1/RC e^-t/CR . (1 – e^-t/CR)+(-) . e^-t/CR(1-/ RC) . e^-t/CR] E^2/2R = (-e^-t/CR/RC + e^-2t/CR/RC + 1/RC. E^-2t/CR) = E^2/2R(2/RC . e^-2t/CR – e^-t/CR/RC) ….(1) For R base max dR/dt = 0 ⇒ 2.e^-t/RC -1 = 0 ⇒ e^-t/CR = ½ ⇒ -t/RC = -In^2 ⇒ t = RC In 2 ∴ Putting t = RC In 2 in equation (1) We get dR/dt = E^2/4R .