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

    11
    ReputationRep:
    So I can't wrap my head around this question, despite it obviously being so simple. I've learnt the 'work done' equation(s):

    Work done (Fs) =
    Fs=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

    At an angle:
    Fs\cos{\theta}=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

    With resistance force:
    Fs\cos{\theta}-Rs=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

    The question is:

    A bicycle of mass 30kg is pushed up a hill inclined at 15 degrees to the horizontal. Calculate the work done in moving the bicycle 70 metres, starting and finishing with the bicycle at rest.

    This 'Work, Energy and Power' section just isn't clicking with me, can someone give me a lil push?
    • Community Assistant
    Offline

    13
    ReputationRep:
    Community Assistant
    (Original post by SamKeene)
    So I can't wrap my head around this question, despite it obviously being so simple. I've learnt the 'work done' equation(s):

    Work done (Fs) =
    Fs=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

    At an angle:
    Fs\cos{\theta}=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

    With resistance force:
    Fs\cos{\theta}-Rs=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

    The question is:

    A bicycle of mass 30kg is pushed up a hill inclined at 15 degrees to the horizontal. Calculate the work done in moving the bicycle 70 metres, starting and finishing with the bicycle at rest.

    This 'Work, Energy and Power' section just isn't clicking with me, can someone give me a lil push?
    Start by drawing a diagram. Then resolve parallel to the plane to get the force acting down the plane. Use Work Done = Force x Distance .
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by SherlockHolmes)
    Start by drawing a diagram. Then resolve parallel to the plane to get the force acting down the plane. Use Work Done = Force x Distance .
    Fs - Rs = 0

    Fs=Rs

     R = \frac{30g}{\cos{75}} Where R is the hypotenuse of the triangle formed.

    But that doesn't work, it should be 30g*cos(75), but why? When resolving up the plane, isn't the amount of resistance of 30g the hypotenuse of the triangle formed?
    • Community Assistant
    Offline

    13
    ReputationRep:
    Community Assistant
    (Original post by SamKeene)
    Fs - Rs = 0

    Fs=Rs

     R = \frac{30g}{\cos{75}} Where R is the hypotenuse of the triangle formed.

    But that doesn't work, it should be 30g*cos(75), but why? When resolving up the plane, isn't the amount of resistance of 30g the hypotenuse of the triangle formed?
    The only force acting down the plane is the parallel component of the weight.

    If you resolve parallel to the plane, you get mgsin(15) which is equal to mgcos(75).

    http://upload.wikimedia.org/wikipedi...e_body.svg.png
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by SherlockHolmes)
    The only force acting down the plane is the parallel component of the weight.

    If you resolve parallel to the plane, you get mgsin(15) which is equal to mgcos(75).
    But \sin{15}=\frac{30g}{R} where R is the slope.

    R\sin{15}=30g

    How come you have mg on the other side?
    • Community Assistant
    Offline

    13
    ReputationRep:
    Community Assistant
    (Original post by SamKeene)
    But \sin{15}=\frac{30g}{R} where R is the slope.

    R\sin{15}=30g

    How come you have mg on the other side?
    Your triangle should have mg as the hypotenuse because you are splitting it into its vertical and horizontal components.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 25, 2015
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
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.