Turn on thread page Beta
    • Thread Starter
    Offline

    7
    ReputationRep:
    A van is pushed from rest with a force of 130N over a distance of 60m. The resistivity forces acting on the van total 90N

    Using the information above how would i calculate the kinetic energy the van gains?
    Offline

    0
    ReputationRep:
    Are you given the mass of the van?
    Offline

    8
    ReputationRep:
    (Original post by JBKProductions)
    A van is pushed from rest with a force of 130N over a distance of 60m. The resistivity forces acting on the van total 90N

    Using the information above how would i calculate the kinetic energy the van gains?
    Work done = force x distance. So if you can work out the total work done by the van over the distance and subtract the work done against resistive forces, you will have the increase in kinetic energy :yep:
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by D-Day)
    Are you given the mass of the van?
    Nope thats all ive been given
    Offline

    0
    ReputationRep:
    (Original post by sonofdot)
    Work done = force x distance. So if you can work out the total work done by the van over the distance and subtract the work done against resistive forces, you will have the increase in kinetic energy :yep:
    Or that.
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by sonofdot)
    Work done = force x distance. So if you can work out the total work done by the van over the distance and subtract the work done against resistive forces, you will have the increase in kinetic energy :yep:
    so is it 2400 J?
    Offline

    0
    ReputationRep:
    2400 J pr 2.4 Kj
    Offline

    8
    ReputationRep:
    (Original post by JBKProductions)
    so is it 2400 J?
    Indeed
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by jjkkll)
    2400 J pr 2.4 Kj
    ok thanks
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by sonofdot)
    Indeed
    if you dont mind me asking another question. The resistance to the motion of a motor boat of mass 780kg is 120N, and the driving force from the propellers is 250N. What distance does it cover while increasing from 2 to 3m/s??
    Offline

    8
    ReputationRep:
    (Original post by JBKProductions)
    if you dont mind me asking another question. The resistance to the motion of a motor boat of mass 780kg is 120N, and the driving force from the propellers is 250N. What distance does it cover while increasing from 2 to 3m/s??
    I would help, but somebody negged me inexplicably on this thread... :mad:

    I'll help really This is like the opposite way around. Find the increase in the kinetic energy (E_k = \frac{1}{2}mv^2) and then the total work done equals the kinetic energy plus the work done against resistance
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by sonofdot)
    I would help, but somebody negged me inexplicably on this thread... :mad:

    I'll help really This is like the opposite way around. Find the increase in the kinetic energy (E_k = \frac{1}{2}mv^2) and then the total work done equals the kinetic energy plus the work done against resistance
    how would i calculate the kinetic energy without the velocity or the distance it travels?
    Offline

    8
    ReputationRep:
    (Original post by JBKProductions)
    how would i calculate the kinetic energy without the velocity or the distance it travels?
    you have the velocity. The gain in kinteic energy = kinetic energy at 3m/s - kinetic energy at 2m/s
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by sonofdot)
    you have the velocity. The gain in kinteic energy = kinetic energy at 3m/s - kinetic energy at 2m/s
    oh ok i see now i didnt understand the question properly i thought i had to work out acc and etc lol but ill try and come bk if i get stuck lol
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by sonofdot)
    you have the velocity. The gain in kinteic energy = kinetic energy at 3m/s - kinetic energy at 2m/s
    so the kinetic energy is 1950J?
    Offline

    0
    ReputationRep:
    (Original post by sonofdot)
    I would help, but somebody negged me inexplicably on this thread... :mad:

    I was not me you crack head, at least ask before negging lol.
    Offline

    8
    ReputationRep:
    (Original post by jjkkll)
    I was not me you crack head, at least ask before negging lol.
    Ok sorry :o: I guess I miscalculated your power... not like my rep is worth anything :rolleyes:
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 14, 2008

3,772

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process

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.