Work, Energy, Power Homework: A Lift of Mass 400kg & 70kg Person

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Homework Help Overview

The problem involves a lift with a mass of 400 kg and a person weighing 70 kg standing on a weighing machine inside the lift. The scenario describes the lift accelerating from rest and reaching a maximum speed after rising 5 meters in 4 seconds. The main question focuses on determining the reading on the weighing machine during the acceleration phase.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of acceleration and question the interpretation of forces acting on the person in the lift. There are attempts to apply kinematic equations and Newton's laws to find the reading on the weighing machine.

Discussion Status

Several participants have provided insights into the forces at play, including the need to consider both the upward force due to acceleration and the downward gravitational force. There is ongoing exploration of how to apply these concepts to find the reading on the scale, with some participants expressing confusion about the relationships between the forces involved.

Contextual Notes

Participants note the lack of explicit information regarding whether the acceleration is uniform, and there are discussions about the total mass of the lift and the person when calculating forces. The urgency of the assignment deadline is also mentioned, indicating time constraints for the original poster.

Klejdi90
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Homework Statement


g= 9.8 Nkg-1

Q. A lift has a mass of 400 kg. A man of mass 70 kg stands on a weighing machine fixed to the floor of the lift. Four seconds after starting from rest the lift has reached its maximum speed and has risen 5 m.

a) What will be the reading on the weighing machine during the point of acceleration?

Homework Equations


A= acceleration
S=distance (5m)
t=Time

A=s/t,F=m*a

The Attempt at a Solution

a = 1.25ms^-1 unsure?

I am really stuck

How would i go about obtaining the change of weight on the scale?
 
Last edited:
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Welcome to PF!

Hi Klejdi90! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)
Klejdi90 said:
a = 1.25ms^-1 unsure?

(ms-1 would be a speed)

no, that would be for a speed of 5, this is a distance …

use one of the other standard constant acceleration equations …

what do you get? :smile:​

How would i go about obtaining the change of weight on the scale?

Use F = ma :wink:
 
?I think this may be it:

S=ut+1/2at2

S=5
U=0
t=4

so 5= 0x4+0.5xax42
a= 5/ 0x4+0.5x42
a=0.625m/s2and for the change in mass

F=ma

so F=75x9.8^-1

mass of man on weighing scale = 75 kg
 
Last edited:
Hi Klejdi90! :smile:
Klejdi90 said:
a=0.625m/s2


Yup! :biggrin:
and for the change in mass

F=ma

so F=75x0.625

which comes to 43.75

There are two forces on the man: mg and N.

N is the same as the weight shown on the weighing machine.

Put them into F = ma, to find N. :smile:
 
tiny-tim said:
Hi Klejdi90! :smile:



Yup! :biggrin:


There are two forces on the man: mg and N.

N is the same as the weight shown on the weighing machine.

Put them into F = ma, to find N. :smile:

sorry to say I'm a little confused :(

F= m*g*?
 
No, "F" in good ol' Newton's "F = ma" is always the total force. :smile:
 
I was wondering if the acceleration was uniform in this question? Why?

Plus would i need to apply the total mass lift and the mans weight so, 400kg+70kg?

There is and upward force due to acceleration and downward force due to g.

So do i do F = ma
F = 70x0.625
F = 43.75N at this point i have the upward force?

The downforce?

Sorry I am very confused I am doing a course which lasts 1 year and teaches A level equivelant physics, plus my teacher is very fast in explaining. I have to give this assignment in tomorrow:(
 
Last edited:
Hi Klejdi90! :smile:
Klejdi90 said:
I was wondering if the acceleration was uniform in this question? Why?

The question doesn't say so (it should), but I expect you're meant to assume so. :wink:
So, F=mga?

erm :redface: … that doesn't even make sense …

how can you multiply two accelerations? :confused:
Plus would i need to apply the total mass lift and the mans weight so, 400kg+70kg?

read the question …
Klejdi90 said:
Q. A lift has a mass of 400 kg. A man of mass 70 kg stands on a weighing machine fixed to the floor of the lift. Four seconds after starting from rest the lift has reached its maximum speed and has risen 5 m.

a) What will be the reading on the weighing machine during the point of acceleration?

what is the reading on the weighing machine equal to? :smile:
 
M=f/a
 
  • #10
Sorry, I meant physically what is the reading on the weighing machine equal to?

(ie, what force does the machine measure? :wink:)
 
  • #11
Kg?
Downforce of man
 
  • #12
That's right … it measures the reaction force between the man and the weighing machine …

by good ol' Newton's third law, the force of the man on the weighing machine is equal (but opposite) to the force of the weighing machine on the man …

so your Ftotal = ma has nothing to do with the lift, does it? :smile:
 

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