A trolley $T$ (mass 5 kg) on a horizontal smooth surface is pulled by a load $L$ (2 kg) through a uniform rope $ABC$ of length 2 m and mass 1kg. As the load falls from $BC = 0$ to $BC = 2\ $m, its acceleration (in $ms^{- 2}$) changes from (Take $g = 10ms^{- 2}$) [image:5_image106.jpeg]

Question

Dr. Abubacker · Accepted Answer

(b) Initially, the weight of load $L$ is the force on the system of mass 8 kg
Acceleration $= \frac{2 	imes 10}{8} = \frac{20}{8}$ units
Towards the end, force $= (2 + 1) 	imes 10$N = 30N
So, acceleration now is $\frac{30}{8}$ units

Q. A trolley $T$ (mass 5 kg) on a horizontal smooth surface is pulled by a load $L$ (2 kg) through a uniform rope $ABC$ of length 2 m and mass 1kg. As the load falls from $BC = 0$ to $BC = 2\ $m, its acceleration (in $ms^{- 2}$) changes from (Take $g = 10ms^{- 2}$)

Explanation

Reviewed by Dr. Abubacker

Want more practice?