Camping equipment weighing 6000 n is pulled across a frozen lake by means of a horizontal rope. the coefficient of kinetic friction is 0.05. how much work is done by the campers in pulling the equipment 1000 m if its speed is increasing at the constant rate of 0.20 m/s2?

I've seen the same image posted on different forums and not explained well so I'm answering this question but in detail.

We know the weight is 6000N and the kinetic friction is .05 . We also know the acceleration is .20 because the rate/velocity is increasing.

Now, you make look at this and be like: Work Formula doesn't involve acceleration!? Neither does the Change in Kinetic Energy Formula!? And to that I respond -- yes. They're not directly involved.

What you have to do, because it's accelerating, is use Newton's Second Law (F=MA).

The forces acting on the sled are friction, weight, and the force applied by the campers.

We can ignore weight because weight is perpendicular to the force applied by the campers.

That leaves us with the force of friction and the force applied.

We can put such forces into F=MA:

Fnet = ma

Fapplied - Ffriction = ma

[subtract friction because it goes against work and force applied]

Now, to solve for m, we take 6000n and divide it by 9.8 to get mass (Weight=mg)

Mass is 612.

Last but surely is friction, we can calculate it by taking the normal force multiplied by our coefficient. Normal force is weight in this case, so we multiply 6000N by .05 .

That should get us: 300

Plug in our values:

Fapplied - 300 = 612(.05)

Fapplied = 612(.05) + 300

Fapplied = 422.4

Now, we multiply by the distance

422.4 (1000)

We get 422400 aka 4.2 x 10^5