30 kg boy in a 1000 kg elevator moving up 5 m/s

  • Thread starter Nick63
  • Start date
  • Tags
    Elevator
In summary: The net external force on a body is the sum of the two forces acting on it, and it depends on the situation.
  • #1
Nick63
13
0

Homework Statement


A 30kg boy is in a 1000kg elevator moving at +5m/s when it comes to a stop in 2 seconds. During the slow down...
1.what is the acceleration of the boy and the elevator?
2. what is the net force acting on the boy?
3.What is the force of the elevator on the boy?

Homework Equations


Acceleration=Net force on object/mass
Force in Newtons=mass*acceleration

The Attempt at a Solution


the boy and elevator have an acceleration of -2.5m/s^2
 
Physics news on Phys.org
  • #2
Nick63 said:

Homework Statement


A 30kg boy is in a 1000kg elevator moving at +5m/s when it comes to a stop in 2 seconds. During the slow down...
1.what is the acceleration of the boy and the elevator?
2. what is the net force acting on the boy?
3.What is the force of the elevator on the boy?

Homework Equations


Acceleration=Net force on object/mass
Force in Newtons=mass*acceleration

The Attempt at a Solution


the boy and elevator have an acceleration of -2.5m/s^2

Okay. What about questions 2 & 3?
 
  • #3
PeroK said:
Okay. What about questions 2 & 3?
For #2 is the two forces on the boy Force of elevator on boy and Force of Earth on Boy(gravity)?
 
  • #4
Nick63 said:
For #2 is the two forces on the boy Force of elevator on boy and Force of Earth on Boy(gravity)?

Yes.
 
  • #5
PeroK said:
Yes.
I think I have the force of Earth on boy at 144 N and the force of the elevator at -2500N, is that right?
 
  • #6
Nick63 said:
I think I have the force of Earth on boy at 144 N and the force of the elevator at -2500N, is that right?

If that were true, the net force on the boy would be ##-2356N##, and hence the boy would accelerate at approximately ##-80m/s^2##.

Do you know the boy's acceleration?
 
  • #7
PeroK said:
If that were true, the net force on the boy would be ##-2356N##, and hence the boy would accelerate at approximately ##-80m/s^2##.

Do you know the boy's acceleration?
I thought it was -2.5m/s^2
 
  • #8
Nick63 said:
I thought it was -2.5m/s^2

It is.

Do you know the boy's mass?
 
  • #9
PeroK said:
It is.

Do you know the boy's mass?
Yes 30kg
 
  • #10
So, you know the mass and the acceleration. What quantity are you trying to find? (For question 2.)
 
  • #11
PeroK said:
So, you know the mass and the acceleration. What quantity are you trying to find? (For question 2.)
the net force on the boy
 
  • #12
How are net force, mass and acceleration related?
 
  • #13
PeroK said:
How are net force, mass and acceleration related?
so 30kg*(-9.8m/s^2+-2.5m/s^2)
 
  • #14
Nick63 said:
so 30kg*(-9.8m/s^2+-2.5m/s^2)

That's not right. You may be confused by what net force means:

Suppose you were stuck in a room where the ceiling was pushing down on you and the floor was pushing up, so that you were getting squashed. In terms of kinematics (the study of motion) there is no net force on you. The two forces cancel out and you do not move. This situation is the same as if no forces were acting on you. That's the nature of forces and motion that you are currently studying. You might describe these as "external forces". There is no net external force in this case.

If you were studying internal stresses and strains on a body, then the two situations would be very different. But, this is not what you are looking for here. You're not measuring how much things are getting squashed, you're measuring how they are moving!

In any case, let me give you the key equation and explain it:

##F = ma##

##F## is the net external force on a body. If two external forces are in opposite directions, then their action tends to cancel out. If you push a body to the right with force ##F_1## and someone else pushes it to the left with force ##F_2##, then ##F = F_1 - F_2##. And, of course, if you both push with the same force ##F_1 = F_2##, then ##F = 0## and the body does not move. (You might be squashing it, but that's not what you're looking for here!)
 
  • #15
PeroK said:
That's not right. You may be confused by what net force means:

Suppose you were stuck in a room where the ceiling was pushing down on you and the floor was pushing up, so that you were getting squashed. In terms of kinematics (the study of motion) there is no net force on you. The two forces cancel out and you do not move. This situation is the same as if no forces were acting on you. That's the nature of forces and motion that you are currently studying. You might describe these as "external forces". There is no net external force in this case.

If you were studying internal stresses and strains on a body, then the two situations would be very different. But, this is not what you are looking for here. You're not measuring how much things are getting squashed, you're measuring how they are moving!

In any case, let me give you the key equation and explain it:

##F = ma##

##F## is the net external force on a body. If two external forces are in opposite directions, then their action tends to cancel out. If you push a body to the right with force ##F_1## and someone else pushes it to the left with force ##F_2##, then ##F = F_1 - F_2##. And, of course, if you both push with the same force ##F_1 = F_2##, then ##F = 0## and the body does not move. (You might be squashing it, but that's not what you're looking for here!)
so then its just 30kg*-2.5m/s^2?
 
  • #16
Nick63 said:
so then its just 30kg*-2.5m/s^2?

Yes.
 

1. How does the weight of the boy affect the elevator's movement?

The weight of the boy does not directly affect the elevator's movement. The elevator's movement is determined by the force of the motor and the counteracting force of gravity. However, the weight of the boy will affect the overall load of the elevator, which may impact the motor's performance.

2. What is the acceleration of the elevator with the boy inside?

The acceleration of the elevator can be calculated using the equation F = ma, where F is the force, m is the mass, and a is the acceleration. In this case, the force is equal to the weight of the elevator and the boy, which is 1030 kg. The acceleration can then be calculated as 1030 kg / 1000 kg = 1.03 m/s².

3. How does the elevator's speed change with the boy's weight?

The elevator's speed is not directly affected by the boy's weight. The speed is determined by the motor's force and the counteracting force of gravity. However, a heavier boy will contribute to a heavier overall load, which may affect the motor's performance and potentially change the speed of the elevator.

4. Does the elevator's speed change if it is moving up or down?

The elevator's speed will change depending on whether it is moving up or down. In this scenario, the elevator is moving up at a speed of 5 m/s. If the elevator were to move down at the same speed, it would experience a negative acceleration due to the counteracting force of gravity, resulting in a slower speed than when moving up.

5. How does the height of the building affect the elevator's movement?

The height of the building does not directly affect the elevator's movement. However, the height of the building may determine the length of time it takes for the elevator to reach its destination, as well as the amount of energy required to move the elevator to a higher or lower floor.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
4K
  • Introductory Physics Homework Help
2
Replies
35
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
23
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
2K
Back
Top