Think about it.
weightapparent=mg+ma where a is up. What if a is down? What if a=-g?
When will your apparent weight be greatest, as measured by a a scale in a moving elevator; when the elevators: (a) accel. downward (b) accel. upward (c) is in free fall (d) moves upward at a constant speed? In what case would you weight be the least? When would it be the same as when you are on the ground?
8 answers
Apparent weight
= m(g+a)
where g=9.8 m/s²
a=acceleration, positive upwards.
(a) W=m(g-|a|)
(b) W=m(g+|a|)
(c) W=m(g-|g|) = 0
(d) W=m(g+|0|) = mg
= m(g+a)
where g=9.8 m/s²
a=acceleration, positive upwards.
(a) W=m(g-|a|)
(b) W=m(g+|a|)
(c) W=m(g-|g|) = 0
(d) W=m(g+|0|) = mg
so the it would be the least on the way down, b/c your force, pushing on the scale is less. And it is greatest on the way up, since your force of push is stronger. And I would say in free fall it is the same as ground state?
Rethink free-fall.
ohh okay so the weight would be the same on the ground if its moving up at a constant speed?
Exactly. When it moves upwards (or downwards) at a uniform velocity, the reaction on the floor is the same as the weight. You have probably experienced this when you ride on an elevator.
sweet!! thank you =)
You're welcome!
1 answer