# Homework Help: Work with a given time period

1. Apr 23, 2010

### tangibleLime

1. The problem statement, all variables and given/known data

a) How much work must you do to push a 13.0 kg block of steel across a steel table ($$\mu_{k}$$=0.6) at a steady speed of 1.30 m/s for 8.30 s?

b) What is your power output while doing so?

2. Relevant equations
$$W=\vec{F}*\Delta S$$
$$F_{f}=u_{k}*n$$
$$F=ma$$

3. The attempt at a solution
First I applied Newton's Second and performed the following:

$$F=ma$$
$$F-F_{f} = ma$$

To find $$F_{f}$$, I used $$F_{f}=u_{k}*n$$ using $$mg$$ for $$n$$ and $$0.6$$ for $$u_{k}$$.

$$F_{f}=u_{k}*mg$$
$$F_{f}=(0.6)*(13)(9.8)$$
$$F_{f}=76.44 N$$

Throwing that back into the NII equation along with substituting the other variables, I got:

$$F-76.44=(13)(9.8)$$
$$F=93.94 N$$

I think I may have done something wrong in that step.

Anyways, if that is correct then this is where I am stuck. I do not know how to apply this information to conclude the work done over the time period supplied.

To solve B, I assume I will simply use the equation $$P=\frac{\Delta E}{\Delta t}$$, where I would use the time supplied for the t value and the answer from part A to the E value.

Any help would be greatly appreciated, thanks!

2. Apr 23, 2010

### ideasrule

How did you get a=9.8? a is 0 because the block is moving at constant speed.

W=Fd. You can calculate F by fixing the mistake I pointed out above, and d is even easier to get.

3. Apr 23, 2010

### jdc15

With your second step, you're overthinking it and making an error by subtracting it from normal force. Force applied is the same as force of friction since it is moving at a constant velocity. This number is correct at 76.44N. 93.94N shouldn't be used.

Now, as for finding work done, as you have listed, W=Fd. You now have force, now you just have to find distance.

For part B your assumption is correct.

4. Apr 23, 2010

### tangibleLime

Ah, thank you!

$$x = x_0 + v_0 t + (1/2) a t^2$$

$$x = 0 + 1.3(8.3) + (1/3)(0)(8.3)^2$$

$$x \approx 825$$

Thanks again!