A light string fixed at one end to a clamp on ground passes over a fixed pulley and hangs at the other side. It makes an angle of `30^(@)` with the ground. A monkey of mass 5 kg climbs up the rope. The clamp can tolerate a vertical force of 40 N only. The maximum acceleration in upward direction with which the monkey can climb safely is:
(neglect friction and take `g=10m//s^(2))`

Answer

`6.00`

Explanation

Let T be the tension in the string.
The upward force exerted on the clamp
`=T sin 30^(@) = T/2`
Given, `T/2 =40 N` or T= 80 N……..(1)
If a is the acceleration of monkey in upward direction.
`a=(T-mg)/m = (80 - 5 xx 10)/5 = 6 m//s^(2)`