A lift is going up. The total mass of the lif | vedclass

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(C) The lift is moving upwards. The tension $T$ in the rope is given by $T = m(g + a)$ when accelerating upwards and $T = m(g - a)$ when decelerating

Vedclass · Accepted Answer

$12000$

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(C) The lift is moving upwards. The tension $T$ in the rope is given by $T = m(g + a)$ when accelerating upwards and $T = m(g - a)$ when decelerating upwards.From the graph,at $t = 11\, s$,the lift is in the interval $10\, s$ to $12\, s$,where it is decelerating.The deceleration $a$ is the slope of the velocity-time graph in this interval: $a = |\frac{v_f - v_i}{t_f - t_i}| = |\frac{0 - 3.6}{12 - 10}| = |\frac{-3.6}{2}| = 1.8\, m/s^2$.Since the lift is moving upwards and decelerating,the equation of motion is $T - mg = -ma$,which gives $T = m(g - a)$.Using $m = 1500\, kg$,$g = 10\, m/s^2$ (standard approximation),and $a = 1.8\, m/s^2$:$T = 1500(10 - 1.8) = 1500 	imes 8.2 = 12300\, N$.If we use $g = 9.8\, m/s^2$:$T = 1500(9.8 - 1.8) = 1500 	imes 8 = 12000\, N$.

$A$ lift is going up. The total mass of the lift and the passenger is $1500\, kg$. The variation in the speed of the lift is as given in the graph. The tension in the rope pulling the lift at $t = 11\, s$ will be ............ $N$.

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