Respuesta :
I will solve this assuming: the rock is lowered at constant speed; the surface is level, and there is no friction
then, the tension in the rope equals the weight of the rock, so T= 50kg x 9.8m/s/s
T=490N; this is the tension in the rope and the force that pulls on the car, so we have
F = ma => a=F/m=490N/1000kg = 0.49m/s/s
if you are initially at rest and accelerate at 0.49m/s/s, it will take a time given by
dist = 1/2 at^2 => t=sqrt[2d/a] to travel 10 m:
t=sqrt[2x10m/0.49m/s/s] = 6.4s which is the time they have to react.
then, the tension in the rope equals the weight of the rock, so T= 50kg x 9.8m/s/s
T=490N; this is the tension in the rope and the force that pulls on the car, so we have
F = ma => a=F/m=490N/1000kg = 0.49m/s/s
if you are initially at rest and accelerate at 0.49m/s/s, it will take a time given by
dist = 1/2 at^2 => t=sqrt[2d/a] to travel 10 m:
t=sqrt[2x10m/0.49m/s/s] = 6.4s which is the time they have to react.
Hello there.
c) How long do the lovers have to apply the brakes before they go over the edge?
6.4s
c) How long do the lovers have to apply the brakes before they go over the edge?
6.4s