People A and B are taking a lift of mass 500 kg which has constant acceleration and the force from the rope that pulls it is 7500 N. The scales where the people stand show a reading of 720 N and 500 N.

Question 1 : Find the acceleration of the lift.Implementing Newton's Second Law we have : F net =Mα lift --->T-F A -F B -Mg =Mα lift ---> α lift =( T-F A -F B -Mg )/M (1)Substituting the relevant values of the forces given by the question we get that α lift = 2.75 m/s 2 ( where g= 9.81 m/s 2 )Question 2 : Find the masses of people A and B.Normally most students would equal the reading of the scale with the multiplication if the mass of the person with gravitational acceleration. This is not the case!! The person also has an upward acceleration equal to the lift's.Again by Newton's second Law we obtain : Person A F A - m A g = m A α lift ---> m A = F A /(g + α lift ) ----> m A = 57.3 kgPerson B F B - m B *g = m A lift ---> m B = F B /(g + α lift ) ----> m B =39.8 kgThis problem although not advanced or complicated can lead to misleading results if students have learned to work mechanically based on solved problems they have seen. As i stated in my description my goal is to avoid that and develop an analytical thinking of the students in order to spot tricky questions like this. Otherwise they would think their solution provided is correct.

GN
Answered by Geri N. Physics tutor

1873 Views

See similar Physics A Level tutors

Related Physics A Level answers

All answers ▸

You are in a vacuum chamber, and you drop a feather and a bowling ball (initially at rest) from a great height. Which will hit the ground first?


A cylindrical specimen of material with diameter 1.5x10^-4 has a breaking stress of 1.3GPa. Calculate the tensile force acting on the specimen at breaking point.


Why are values for gravitational potential always negative?


What's the difference between Potential Difference and Electromotive Force


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning