# Homework Help: How much work must you do to push a 13 block of steel across a steel table at a ste

1. Nov 24, 2009

### bjbaldwi

How much work must you do to push a 13g block of steel across a steel table(coefficent of friction=.6) at a steady speed of 1.2 for 8.8 ?

I know that Work=Force*Distance

i know what the D is but i dont know how to calculate the force.
i tried F=MA and got Fnet= Force pushed -Friction force
i really have no i dea tho. I need help please.

2. Nov 24, 2009

### Staff: Mentor

Re: How much work must you do to push a 13 block of steel across a steel table at a

How do you calculate the friction force?

3. Nov 24, 2009

### bjbaldwi

Re: How much work must you do to push a 13 block of steel across a steel table at a

Steady speed says the friction is constant and friction force is kinetic friction*normal force or Fk*mg mg= normal force i think

4. Nov 24, 2009

### Staff: Mentor

Re: How much work must you do to push a 13 block of steel across a steel table at a

OK, good. But what does Fnet equal? (What's the acceleration of the block?)

5. Nov 24, 2009

### bjbaldwi

Re: How much work must you do to push a 13 block of steel across a steel table at a

sorry its 13 kg 1.2 m/s and 8.8 s

6. Nov 24, 2009

### bjbaldwi

Re: How much work must you do to push a 13 block of steel across a steel table at a

the acceleration is zero because it is a constant velocity

7. Nov 24, 2009

### Staff: Mentor

Re: How much work must you do to push a 13 block of steel across a steel table at a

Good. So what's Fnet? And what does that tell you about Fpush?

8. Nov 24, 2009

### bjbaldwi

Re: How much work must you do to push a 13 block of steel across a steel table at a

Fnet is equal to zero since the acc. is zero. F push is equal to friction force?

9. Nov 24, 2009

### Staff: Mentor

Re: How much work must you do to push a 13 block of steel across a steel table at a

Exactly! You should be able to calculate the work now.

10. Nov 24, 2009

### bjbaldwi

Re: How much work must you do to push a 13 block of steel across a steel table at a

Fpush is equal to 0 and f net is equal to friction force

so Fnet=Fpush+Ffriction
Fpush=0
Fnet=Ffriction
Ffriction=Fnet =(.6)mg---->.6*13*9.8=76.44
so
Work=Fnet*D
D = 1.2m/s*8.8s---->10.56 m
so
76.44*10.56=807.2 J

11. Nov 24, 2009

### bjbaldwi

Re: How much work must you do to push a 13 block of steel across a steel table at a

Thanks for all your help : )

12. Nov 24, 2009

### Staff: Mentor

Re: How much work must you do to push a 13 block of steel across a steel table at a

No, you had it right before. Fnet = 0 so Fpush = Ffriction.

No, but that doesn't affect your calculation, which is correct.

It's the force Fpush that is doing the work.