Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    14
    ReputationRep:
    A particle of mass 2m can move on the rough surface of a plane inclined at an angle theta to the horizontal where sintheta=3/5 . A second particle of mass 3m hangs freely attached to a light inextensible string which passes over a smooth pulley fixed at D. The other end of the string is attached to the 2m mass on the inclined plane. The coefficien of friction is 0.25. The system is released from rest with the string taut and the 2m mass moves up a line of greatest slope of the plane. When each particle has moved a distance s, the 2m mass has not reached the pulley and the 3m mass has not reached the ground.

    A) find an expression for the potential lost by the system when each particle has moved a distance of s.
    Energy lost=9mgs/5
    B) When each particle has moved a distance of s, they are moving at a speed v. Find an expression for v2 in terms of s.
    I have found so far:
    a=2.94 ms^-2
    Friction= 3mg/10

    What do I do now?
    • Study Helper
    Offline

    9
    ReputationRep:
    Study Helper
    (Original post by bobbricks)
    A particle of mass 2m can move on the rough surface of a plane inclined at an angle theta to the horizontal where sintheta=3/5 . A second particle of mass 3m hangs freely attached to a light inextensible string which passes over a smooth pulley fixed at D. The other end of the string is attached to the 2m mass on the inclined plane. The coefficien of friction is 0.25. The system is released from rest with the string taut and the 2m mass moves up a line of greatest slope of the plane. When each particle has moved a distance s, the 2m mass has not reached the pulley and the 3m mass has not reached the ground.

    A) find an expression for the potential lost by the system when each particle has moved a distance of s.
    Energy lost=9mgs/5
    B) When each particle has moved a distance of s, they are moving at a speed v. Find an expression for v2 in terms of s.
    I have found so far:
    a=2.94 ms^-2
    Friction= 3mg/10

    What do I do now?
    For (B) Use gain of kinetic energy=loss of potential energy - work done againbst friction.
    i.e. the work-energy principle
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by brianeverit)
    For (B) Use gain of kinetic energy=loss of potential energy - work done againbst friction.
    i.e. the work-energy principle
    I did 1/2 (2m)v^2=9(2m)gs/5 -2(2m)gs/5
    but that doesn't give the correct answer??

    EDIT: I got v^2 =14gs/5 which I think is correct..?
    • Study Helper
    Offline

    9
    ReputationRep:
    Study Helper
    (Original post by bobbricks)
    I did 1/2 (2m)v^2=9(2m)gs/5 -2(2m)gs/5
    but that doesn't give the correct answer??

    EDIT: I got v^2 =14gs/5 which I think is correct..?
    what about the gain in k.e. of the mass 3m?
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 8, 2015
Poll
Do I go to The Streets tomorrow night?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.