Calculating Work & Forces: A Toboggan Problem Solved in 2 Hours

[tongtong]

I've been stuck on this problem for two hours and my teacher hasn't taught properly.

A parent pulls a toboggan with three children at a constant velocity for 38 m along a horizontal snow-covered trail. The total mass of the children and the toboggan is 66 kg. The force the parent exerts is 58 N [18 degrees above the horizontal].
a) Determine the magnitude of the normal force and the coefficient of the kinetic fiction.
b) Determine the work done on the toboggan by kniectic friction.
c) List three forces, or components of forces, that do zero work on the toboggan.
d) How much work does the parent do on the toboggan in pulling it the first 25m?

This is what I have gotten so far:
a) Fn = mg = 66*9.8 N
c) Fn, Fg, Fa of the children on the tobaggan
d) W = F*d*cos theta = 58*38*cos 18 = 2096.13 N
Is the above correct?

For part a, I don't understand how to solve for the coefficient. I only know of the formula, uk = Fk / Fn, but I don't think I have Fk?

Thanks for the help in advance, much appreciated.

 

[thermodynamicaldude]

A constant velocity means that the forces on the sled are balanced. (Newton's first law) That thus means that the force the parents exert on the sled = frictional force pushing the sled back. Thus,

Fk(frictional force) = force the parents exert on sled (or 58 * cos18)

 

[M98Ranger]

First of all, I understand your frustration with being stuck on a problem for such a long time. It can be frustrating when you feel like you haven't been taught the necessary concepts to solve a problem. However, it's important to remember that problem-solving and critical thinking are important skills to develop in any subject, and sometimes it takes time and effort to figure things out on your own. With that being said, let's take a look at the problem and see if we can help you out.

a) To solve for the coefficient of kinetic friction, we need to use the formula uk = Fk / Fn. In this case, we have been given the force the parent exerts (58 N) and the normal force (Fn = mg = 66*9.8 N). So we can plug those values into the formula and solve for uk. This gives us uk = 58 / (66*9.8) = 0.0885.

b) To determine the work done on the toboggan by kinetic friction, we can use the formula W = Fd = Fk *d. We have already calculated the coefficient of kinetic friction (uk = 0.0885) and we know the distance the toboggan was pulled (d = 38 m). So we can plug those values into the formula and solve for Fk. This gives us Fk = uk * Fn = 0.0885 * (66*9.8) = 58.4 N. Therefore, the work done on the toboggan by kinetic friction is W = Fk * d = 58.4 * 38 = 2219.2 J.

c) The three forces that do zero work on the toboggan are the normal force (Fn), the force of gravity (Fg), and the applied force (Fa) of the children on the toboggan. This is because these forces are perpendicular to the direction of motion of the toboggan and therefore do not contribute to the work being done.

d) To calculate the work done by the parent on the toboggan in pulling it the first 25 m, we can use the same formula as in part b. So W = Fd = F * d * cos theta. We already know the force exerted by the parent (F = 58 N) and the distance pulled (d = 25 m). We also know the angle between the