The pull of cable is 5.24x10^3N
The force exerted by the man =75N
Is it correct?
An elevator with a man inside, of gross mass 500KG, starts moving upwards with a costant acceleration and acquires a velocity os 2m/s after travelling a distance os 3m.Find the pull in the cables during the accelerated motion.
If the elevator, before stopping, moves with a constant deceleration from a constant velocity of 2m/s and comes to rest in 2 second, calculate the force exerted by the man of 75kg mass on the floor on the elevator.
2 answers
first, find the acceleration.
Vf^2=Vi^2+2ad
4=0+2a*3
a=2/3 m/s^2
then
force= m(g+a)=500(9.8+2/3) =
Vf^2=Vi^2+2ad
4=0+2a*3
a=2/3 m/s^2
then
force= m(g+a)=500(9.8+2/3) =
1 answer