A monkey coming down along a vertical light rope with constant velocity of "4m/" sec.Find the tension in rope due to this.Mass of monkey is "6kg (Take `g=10m/sec^(2`

A 1kg stone at the end of 1m long string is whirled in a vertical circle at a constant speed of 4m//s . The tension in the string is 6N , when the stone is at (g=10m//s^(2))

A 20 kg monkey slides down a vertical rope with a constant acceleration of 7 ms^(-2) . If g = 10 m//s^(-2) , what is the tension in the rope ?

A ship is moving due east with a velocity of 12 m//sec , a truck is moving across on the ship with velocity 4m//sec .A monkey is climbing the vertical pole mounted on the truck with a velocity of 3m//sec .Find the velocity of the monkey as observed by the man on the shore (m//sec)

A 20 kg bucket is lowered by a rope with constant velocity of 0.5m/s. What is the tension in the rope? A 20kg bucket is lowered with a constant downward acceleration of 1m//s^(2) . What is the tension in the rope? A 10kg bucket is raised with a constant upward acceleration with same magnitude a.

Two monkey of masses 10 kg and 8 kg are moving along a verticle rope as shown in fig. the former climbing up with an acceleration of 2 ms^(-2) , while the later coming down with a uniform velocity of 2 ms^(-2) . Find the tension in the rope at the fixed support.

A stone of mass 1 kg is tied to a string 2m long and it's rotated at constant speed of 40 ms^(-1) in a vertical circle. The ratio of the tension at the top and the bottom is [Take g=10m s^(2) ]

In order to raise a mass of 100kg a of mass 60kg fastens a rope to it and passes the rope over a smooth pulley. He climbs the rope with acceleration 5g//4 relative to the rope. The tension in the rope is: Take g=10m//s^(2)

Mass of a person sitting in a lift is 50 kg . If lift is coming down with a constant acceleration of 10 m//sec^(2) . Then the reading of spring balance will be (g=10m//sec^(2))