A 72.2-kg person stands on a scale in an elevator. What is the apparent weight when the elevator is (a) accelerating upward with an acceleration of 1.53 m/s2, (b) moving upward at a constant speed, and (c) accelerating downward with an acceleration of 1.44 m/s2?
a) weight in N = 72.2 x (9.8+1.53) = 818N
b) weight in N = 72.2 x 9.8 = 708N
c) weight in N = 72.2 x (9.8-1.44) = 604N
you multiply the mass by gravity +/- the acceleration depending on if the elevator is moving upwards or downwards.
(a) Well, if the elevator is going up and accelerating, it's like the scale is playing a little prank on our 72.2-kg person. The apparent weight will be more than their actual weight since those upward forces are adding up. So, grab a seatbelt because their apparent weight will be a bit heavier. Can't let them float away, after all.
(b) Now, when the elevator is moving upward at a constant speed, there's no acceleration going on. It's like a smooth ride on a rollercoaster with no loops. The scale is like, "Hey, 72.2-kg person, just chill. No tricks this time." So, their apparent weight will be exactly equal to their actual weight. No surprises, just good old physics behaving.
(c) Hold on tight! When the elevator is accelerating downward, it's like gravity is getting stronger. It's like being on a really intense ride at an amusement park, but without the cotton candy. The scale is like, "Hey, 72.2-kg person, how about we pull a little prank and make your apparent weight lighter?" So, their apparent weight will be a bit lighter than their actual weight.
Remember, these are all circus tricks played by physics! It's all in good fun, as long as you don't drop any pies on your head. Enjoy the ride!
To find the apparent weight of a person in different situations, we need to consider the forces acting on the person. The apparent weight is the force exerted by the scale on the person.
(a) When the elevator is accelerating upward with an acceleration of 1.53 m/s^2:
The net force acting on the person can be calculated using Newton's second law:
F_net = m * a
F_net = 72.2 kg * 1.53 m/s^2
F_net = 110.646 N
The apparent weight is equal to the magnitude of the net force, so the apparent weight is 110.646 N.
(b) When the elevator is moving upward at a constant speed:
When the elevator is moving at a constant speed, there is no acceleration. Therefore, the net force on the person is zero since there are no unbalanced forces. The apparent weight is equal to the person's actual weight, which is the force due to gravity.
The weight of the person is given by:
F_weight = m * g
F_weight = 72.2 kg * 9.8 m/s^2
F_weight = 708.76 N
The apparent weight is 708.76 N.
(c) When the elevator is accelerating downward with an acceleration of 1.44 m/s^2:
Similar to the first case, the net force on the person can be calculated using Newton's second law:
F_net = m * a
F_net = 72.2 kg * -1.44 m/s^2 (Note: we have taken the acceleration as negative because it is directed downward)
F_net = -104.128 N
The apparent weight is the magnitude of the net force, so the apparent weight is 104.128 N.
To summarize:
(a) The apparent weight when the elevator is accelerating upward is 110.646 N.
(b) The apparent weight when the elevator is moving upward at a constant speed is 708.76 N.
(c) The apparent weight when the elevator is accelerating downward is 104.128 N.
To find the apparent weight of a person in different situations in the elevator, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is equal to the weight of the person.
Apparent weight is defined as the force exerted on an object by a scale when the object is in contact with the scale. In other words, it is the reading on the scale.
Let's calculate the apparent weight in each situation:
(a) Accelerating upward with an acceleration of 1.53 m/s^2:
In this case, the net force acting on the person is the sum of their actual weight (mg) and the force due to the upward acceleration (ma). The apparent weight is given by the formula F_net = m * (g + a).
m = 72.2 kg (mass of the person)
g = 9.8 m/s^2 (acceleration due to gravity)
a = 1.53 m/s^2 (acceleration)
F_net = 72.2 kg * (9.8 m/s^2 + 1.53 m/s^2)
F_net = 72.2 kg * 11.33 m/s^2
F_net = 818.726 N
Therefore, the apparent weight when the elevator is accelerating upward is 818.726 N.
(b) Moving upward at a constant speed:
When the elevator is moving at a constant speed, the net force acting on the person is zero because the acceleration is zero. Hence, the apparent weight is equal to the actual weight.
F_net = m * g
F_net = 72.2 kg * 9.8 m/s^2
F_net = 708.76 N
Therefore, the apparent weight when the elevator is moving upward at a constant speed is 708.76 N.
(c) Accelerating downward with an acceleration of 1.44 m/s^2:
Similar to case (a), the net force is the sum of the person's actual weight and the force due to the downward acceleration. The formula for the apparent weight becomes F_net = m * (g - a).
m = 72.2 kg (mass of the person)
g = 9.8 m/s^2 (acceleration due to gravity)
a = 1.44 m/s^2 (acceleration)
F_net = 72.2 kg * (9.8 m/s^2 - 1.44 m/s^2)
F_net = 72.2 kg * 8.36 m/s^2
F_net = 604.392 N
Therefore, the apparent weight when the elevator is accelerating downward is 604.392 N.