A question regarding work with friction

[physicskid72]

Homework Statement

an object having a mass of 25 kg is moving at a steady speed on ice. The coefficient of friction is 0.12.

a.) how much work is done moving the object 35 m?

b.) how much work is done if the object is accelerated at 0.55 m/s/s over a distance of 35 m?

Homework Equations

W=Fd Ff= mewFn F=ma

The Attempt at a Solution

a.)W= 0.12(25)(9.80)(35)
= 1029 J

b.) F= ma
= 25(0.55)
= 13.75 J
This is where i am confused, do i add up the work from part a to part be to get the total work? thanks

 

[Doc Al]

a.)W= 0.12(25)(9.80)(35)
= 1029 J

OK. This is the work done by the applied force, which you know must equal the force of friciton.

b.) F= ma
= 25(0.55)
= 13.75 J

You found the net force, not any work done. In order to achieve the given acceleration, what must the applied force be? What work does it do?

 

[kuruman]

In part (a) the work done by the net force is zero because the net force is zero. You know that the net force is zero because the object moves at a steady speed. That is why you can say that the work done by the pulling force is the opposite to the "work done by friction."

In part (b) the work done done by the net force is not zero because the object accelerates. The "work done by friction" is the same as in part (a), but the work done by the pulling force is not. So you need that to say that

(Work done by pulling force) - (work done by friction) = Work done by net force

BTW, 25(0.55) is just mass times acceleration. That is the net force expressed in Newtons. It cannot be a work expressed in Joules.

 

[Doc Al]

In part (a) the work done by the net force is zero because the net force is zero. You know that the net force is zero because the object moves at a steady speed. That is why you can say that the work done by the pulling force is the opposite to the "work done by friction."

I doubt they mean the net work done by all forces, which is of course zero. It requires work (by the applied force) to move that mass--that work is what they want you to calculate. (At least that's how I see it. It is a bit ambiguously worded though, so your interpretation may well be correct.)

 

[physicskid72]

ok so i see i forgot to multiply the netforce by the distance so...
f=ma
= 25(0.55)
= 13.75

W=Fd
= (13.75)(35)
= 481.25 J

ok so the work done by the net force is 481.25 J - and that means that the because the object is now accelerating this external force must be doing 481.25 J more work than the force of friction in the opposite direction.

so... W= 1029+481.25
= 1510.25 J

Is this correct?

 

[Doc Al]

ok so i see i forgot to multiply the netforce by the distance so...
f=ma
= 25(0.55)
= 13.75

W=Fd
= (13.75)(35)
= 481.25 J

ok so the work done by the net force is 481.25 J - and that means that the because the object is now accelerating this external force must be doing 481.25 J more work than the force of friction in the opposite direction.

so... W= 1029+481.25
= 1510.25 J

Yes, that's essentially correct. But I would express it a bit differently. I would calculate the applied force, then the work done by that force. The applied force can be found using Newton's 2nd law: F_applied - μmg = ma.

(Of course, you'll get the same answer as you did. )