University of Oxford Decisions Day, join the Official Offer Holder thread here>>

Nominations for the TSR Community Awards 2024 OPEN NOW!

University of Oxford Decisions Day, join the Official Offer Holder thread here>>

Nominations for the TSR Community Awards 2024 OPEN NOW!

This discussion is now closed.

Check out other Related discussions

What is meaning of the work done against the resistance?

The total work done? The work done by the resistance?
Or something else??

What?

superkillball

The total work done? The work done by the resistance?
Or something else??

I'm also confused! Do you mean that if you had a cyclist peddling up a hill, how would you calculate the work down using the work-energy principle?!

OP

Hoofbeat

I'm also confused! Do you mean that if you had a cyclist peddling up a hill, how would you calculate the work down using the work-energy principle?!

Exactly

superkillball

Exactly

if you have a cyclist going up a hill, the work done is the work done in going upwards and the work done against friction

Force times distance...

mik1a

Force times distance...

...equals work done....

Katie Heskins

...equals work done....

..against the force...

mik1a

Force times distance...

In the direction of the force.

mik1a

Force times distance...

just applying that though doesnt take account of the steepness of the hill nor the friction

Yes it does, because the force will have to be larger to move the body the same distance.

if you want the vector defenition,

W = -∫F.dr

mik1a

Yes it does, because the force will have to be larger to move the body the same distance.

good point...so in which case...f=
the frictional force
added to the force that is required to move the mass upwards. could you get this using f=ma

elpaw

if you want the vector defenition,

W = -∫F.dr

lol cheers

If you are talking about a cyclist peddling up a hill, there is NO work done against friction between the slope and the cyclist because the wheels do not slide but rotate.

The energy used is converted to the gain in gravitational potential energy of the cyclist, the rotational kinetic energy of the wheels, work done against friction between the chain, the pulley and the wheels of the cyclist (and translational kinetic energy of the cyclist if it is not at rest finally).

If there is no work done against friction between the chain, the pulley and the wheels of the cyclist, the force required is equal to the component of its weight along the slope plus the friction. The additional work done does not go to overcome the friction, but to the rotational kinetic energy of the wheels.

keisiuho

If you are talking about a cyclist peddling up a hill, there is NO work done against friction between the slope and the cyclist because the wheels do not slide but rotate.

but we normally model the cyclyst as a particle, so there is friction, and there is no rotational kinetic energy (which isnt covered till M4)

Latest

Trending

Articles for you

A-level Economics Paper 1 (Edexcel) unofficial markscheme - 15 May 2024

How to write an excellent personal statement in 10 steps

How to write an excellent personal statement in 10 steps

How to write your law personal statement...and make it really stand out

How to write your law personal statement...and make it really stand out

Can you speak lawyer?

Can you speak lawyer?

Try out the app

Continue on web